// Numbas version: finer_feedback_settings {"name": "Core Maths Skills Final Exam Practice Questions", "metadata": {"description": "

These practice questions can help you revise basic mathematical concepts needed to succeed when studying STEM programs.

", "licence": "None specified"}, "duration": 0, "percentPass": "0", "showQuestionGroupNames": true, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Core Maths Skills", "pickingStrategy": "all-shuffled", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], []], "questions": [{"name": "Solve equations which include a single even power (e.g. x^even=blah)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "tags": [], "metadata": {"description": "

Questions to test if the student knows the inverse of an even power (and how to solve equations that contain a single power that is even).

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Please complete the following.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"cxcoeff": {"name": "cxcoeff", "group": "c", "definition": "if(cc-cb>0,random(2..12),random(-12..-2))", "description": "", "templateType": "anything", "can_override": false}, "intrhs": {"name": "intrhs", "group": "a", "definition": "intsoln^intpower\n", "description": "", "templateType": "anything", "can_override": false}, "bpower": {"name": "bpower", "group": "b", "definition": "powers[1]", "description": "", "templateType": "anything", "can_override": false}, "db": {"name": "db", "group": "Ungrouped variables", "definition": "random(-100..100 except -1..1)", "description": "", "templateType": "anything", "can_override": false}, "bc": {"name": "bc", "group": "b", "definition": "bsoln^bpower*bxcoeff+bb", "description": "", "templateType": "anything", "can_override": false}, "ddenom": {"name": "ddenom", "group": "Ungrouped variables", "definition": "random(2..15)", "description": "", "templateType": "anything", "can_override": false}, "powers": {"name": "powers", "group": "a", "definition": "shuffle([2,4,6,8])", "description": "", "templateType": "anything", "can_override": false}, "bb": {"name": "bb", "group": "b", "definition": "random(1..100)", "description": "", "templateType": "anything", "can_override": false}, "dc": {"name": "dc", "group": "Ungrouped variables", "definition": "random(2..100)", "description": "", "templateType": "anything", "can_override": false}, "cc": {"name": "cc", "group": "c", "definition": "random(-100..100)", "description": "", "templateType": "anything", "can_override": false}, "brhs": {"name": "brhs", "group": "b", "definition": "(bc-bb)/bxcoeff", "description": "", "templateType": "anything", "can_override": false}, "cb": {"name": "cb", "group": "c", "definition": "random(-100..100 except [0,cc])", "description": "", "templateType": "anything", "can_override": false}, "bxcoeff": {"name": "bxcoeff", "group": "b", "definition": "random(-3..3 except 0..1)", "description": "", "templateType": "anything", "can_override": false}, "intsoln": {"name": "intsoln", "group": "a", "definition": "switch(intpower=2, random(2..12), intpower=4, random(2..5), intpower=6, random(2..3), 2)", "description": "", "templateType": "anything", "can_override": false}, "intpower": {"name": "intpower", "group": "a", "definition": "powers[0]", "description": "", "templateType": "anything", "can_override": false}, "bsoln": {"name": "bsoln", "group": "b", "definition": "switch(bpower=2, random(2..10), bpower=4, random(2..4), bpower=6, random(2..3), 2)", "description": "", "templateType": "anything", "can_override": false}, "cpower": {"name": "cpower", "group": "c", "definition": "powers[2]", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["dc", "db", "ddenom"], "variable_groups": [{"name": "a", "variables": ["powers", "intpower", "intrhs", "intsoln"]}, {"name": "b", "variables": ["bpower", "bsoln", "bxcoeff", "bb", "bc", "brhs"]}, {"name": "c", "variables": ["cpower", "cc", "cb", "cxcoeff"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "marked_original_order (Mark the gaps in the original order, mainly to establish if every gap has a valid answer):\n map(\n mark_part(gap[\"path\"],studentAnswer),\n [gap,studentAnswer],\n zip(gaps,studentAnswer)\n )\n\ninterpreted_answers (The interpreted answers for each gap, in the original order):\n map(\n res[\"values\"][\"interpreted_answer\"],\n res,\n marked_original_order\n )\n\nanswers (The student's answers to each gap): interpreted_answers\n\ngap_feedback (Feedback on each of the gaps):\n map(\n try(\n let(\n answer, studentAnswer[answer_number],\n result, submit_part(gaps[gap_number][\"path\"],answer),\n gap, gaps[gap_number],\n name, gap[\"name\"],\n noFeedbackIcon, not gap[\"settings\"][\"showFeedbackIcon\"],\n non_warning_feedback, filter(x[\"op\"]<>\"warning\",x,result[\"feedback\"]),\n assert(noFeedbackIcon,\n assert(name=\"\" or len(gaps)=1 or len(non_warning_feedback)=0,feedback(translate('part.gapfill.feedback header',[\"name\": name])))\n );\n concat_feedback(non_warning_feedback, if(marks>0,result[\"marks\"]/marks,1/len(gaps)), noFeedbackIcon);\n result\n ),\n err,\n fail(translate(\"part.gapfill.error marking gap\",[\"name\": gaps[gap_number][\"name\"], \"message\": err]))\n ),\n [gap_number, answer_number],\n zip(gap_adaptive_order, gap_adaptive_order)\n )\n\nall_valid (Are the answers to all of the gaps valid?):\n all(map(res[\"valid\"], res, marked_original_order))\n\ncheck_answers:\n if(\n answers[1]=answers[0],\n sub_credit(1/2, \"Numbers cannot be the same. \"),\n feedback(\"\")\n )\n\nmark:\n assert(all_valid or not settings[\"sortAnswers\"], fail(translate(\"question.can not submit\")));\n apply(answers);\n apply(check_answers);\n apply(gap_feedback)\n\ninterpreted_answer:\n answers\n\npre_submit:\n map(\n let(\n answer, studentAnswer[answer_number],\n result, submit_part(gaps[gap_number][\"path\"],answer),\n check_pre_submit(gaps[gap_number][\"path\"], answer, exec_path)\n ),\n [gap_number,answer_number],\n zip(gap_order,answer_order)\n )", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

If $x^\\var{intpower}=\\var{intrhs}$, then $x=$ [[0]], or [[1]].

", "stepsPenalty": "2", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Note: Suppose you wanted to solve $x^2=4$, that is, you wanted the number that squares to give four. The number $2$ comes to mind, however, $-2$ is also a solution since $(-2)^2=(-2)\\times(-2)=4$. Recall the product of two negatives is a positive, so the product of any even number of negative numbers is positive. This means when we find a positive solution to an equation like $x^4=10\\,000$ there will also be a negative solution, in particular, the solution to $x^4=10\\,000$ would be $x=\\pm\\sqrt[4]{10\\,000}=\\pm10$.

a) Since the power in $x^\\var{intpower}=\\var{intrhs}$ is even we will take the plus or minus $\\var{intpower}$nd rd th root to get two solutions.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$x^\\var{intpower}$	$=$	$\\var{intrhs}$

$\\sqrt[\\var{intpower}]{x^\\var{intpower}}$	$=$	$\\pm\\sqrt[\\var{intpower}]{\\var{intrhs}}$

$x$	$=$	$\\pm\\var{intsoln}$

That is, $x$ equals $-\\var{intsoln}$ or $\\var{intsoln}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "{intsoln}", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "{-intsoln}", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "{-intsoln}", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "{intsoln}", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "sortAnswers": true}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "marked_original_order (Mark the gaps in the original order, mainly to establish if every gap has a valid answer):\n map(\n mark_part(gap[\"path\"],studentAnswer),\n [gap,studentAnswer],\n zip(gaps,studentAnswer)\n )\n\ninterpreted_answers (The interpreted answers for each gap, in the original order):\n map(\n res[\"values\"][\"interpreted_answer\"],\n res,\n marked_original_order\n )\n\nanswers (The student's answers to each gap): interpreted_answers\n\ngap_feedback (Feedback on each of the gaps):\n map(\n try(\n let(\n answer, studentAnswer[answer_number],\n result, submit_part(gaps[gap_number][\"path\"],answer),\n gap, gaps[gap_number],\n name, gap[\"name\"],\n noFeedbackIcon, not gap[\"settings\"][\"showFeedbackIcon\"],\n non_warning_feedback, filter(x[\"op\"]<>\"warning\",x,result[\"feedback\"]),\n assert(noFeedbackIcon,\n assert(name=\"\" or len(gaps)=1 or len(non_warning_feedback)=0,feedback(translate('part.gapfill.feedback header',[\"name\": name])))\n );\n concat_feedback(non_warning_feedback, if(marks>0,result[\"marks\"]/marks,1/len(gaps)), noFeedbackIcon);\n result\n ),\n err,\n fail(translate(\"part.gapfill.error marking gap\",[\"name\": gaps[gap_number][\"name\"], \"message\": err]))\n ),\n [gap_number, answer_number],\n zip(gap_adaptive_order, gap_adaptive_order)\n )\n\nall_valid (Are the answers to all of the gaps valid?):\n all(map(res[\"valid\"], res, marked_original_order))\n\ncheck_answers:\n if(\n answers[1]=answers[0],\n sub_credit(1/2, \"Numbers cannot be the same. \"),\n feedback(\"\")\n )\n\nmark:\n assert(all_valid or not settings[\"sortAnswers\"], fail(translate(\"question.can not submit\")));\n apply(answers);\n apply(check_answers);\n apply(gap_feedback)\n\ninterpreted_answer:\n answers\n\npre_submit:\n map(\n let(\n answer, studentAnswer[answer_number],\n result, submit_part(gaps[gap_number][\"path\"],answer),\n check_pre_submit(gaps[gap_number][\"path\"], answer, exec_path)\n ),\n [gap_number,answer_number],\n zip(gap_order,answer_order)\n )", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

If $\\simplify{{bxcoeff}y^{bpower}+{bb}}=\\var{bc}$, then $y=$ [[0]], or [[1]].

Given $\\simplify{{bxcoeff}y^{bpower}+{bb}}=\\var{bc}$, we can rearrange the equation to get $y^\\var{bpower}$ by itself and then we can take the plus or minus $\\var{bpower}$nd rd th root to get $y$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{{bxcoeff}y^{bpower}+{bb}}$	$=$	$\\var{bc}$

$\\simplify{{bxcoeff}y^{bpower}}$	$=$	$\\simplify[basic]{{bc}-{bb}}$

$\\simplify{{bxcoeff}y^{bpower}}$	$=$	$\\simplify{{bc-bb}}$

$y^\\var{bpower}$	$=$	$\\simplify[!basic]{{bc-bb}/{bxcoeff}}$

$y^\\var{bpower}$	$=$	$\\simplify{{bc-bb}/{bxcoeff}}$

$\\sqrt[\\var{bpower}]{y^\\var{bpower}}$	$=$	$\\pm\\sqrt[\\var{bpower}]{\\var{brhs}}$

$y$	$=$	$\\pm\\var{bsoln}$

That is, $y$ equals $-\\var{bsoln}$, or $\\var{bsoln}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "{bsoln}", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "{-bsoln}", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "{-bsoln}", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "{bsoln}", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "sortAnswers": true}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "marked_original_order (Mark the gaps in the original order, mainly to establish if every gap has a valid answer):\n map(\n mark_part(gap[\"path\"],studentAnswer),\n [gap,studentAnswer],\n zip(gaps,studentAnswer)\n )\n\ninterpreted_answers (The interpreted answers for each gap, in the original order):\n map(\n res[\"values\"][\"interpreted_answer\"],\n res,\n marked_original_order\n )\n\nanswers (The student's answers to each gap): interpreted_answers\n\ngap_feedback (Feedback on each of the gaps):\n map(\n try(\n let(\n answer, studentAnswer[answer_number],\n result, submit_part(gaps[gap_number][\"path\"],answer),\n gap, gaps[gap_number],\n name, gap[\"name\"],\n noFeedbackIcon, not gap[\"settings\"][\"showFeedbackIcon\"],\n non_warning_feedback, filter(x[\"op\"]<>\"warning\",x,result[\"feedback\"]),\n assert(noFeedbackIcon,\n assert(name=\"\" or len(gaps)=1 or len(non_warning_feedback)=0,feedback(translate('part.gapfill.feedback header',[\"name\": name])))\n );\n concat_feedback(non_warning_feedback, if(marks>0,result[\"marks\"]/marks,1/len(gaps)), noFeedbackIcon);\n result\n ),\n err,\n fail(translate(\"part.gapfill.error marking gap\",[\"name\": gaps[gap_number][\"name\"], \"message\": err]))\n ),\n [gap_number, answer_number],\n zip(gap_adaptive_order, gap_adaptive_order)\n )\n\nall_valid (Are the answers to all of the gaps valid?):\n all(map(res[\"valid\"], res, marked_original_order))\n\ncheck_answers:\n if(\n answers[1]=answers[0],\n sub_credit(1/2, \"Numbers cannot be the same. \"),\n feedback(\"\")\n )\n\nmark:\n assert(all_valid or not settings[\"sortAnswers\"], fail(translate(\"question.can not submit\")));\n apply(answers);\n apply(check_answers);\n apply(gap_feedback)\n\ninterpreted_answer:\n answers\n\npre_submit:\n map(\n let(\n answer, studentAnswer[answer_number],\n result, submit_part(gaps[gap_number][\"path\"],answer),\n check_pre_submit(gaps[gap_number][\"path\"], answer, exec_path)\n ),\n [gap_number,answer_number],\n zip(gap_order,answer_order)\n )", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

For this question input the exact value by using a fractional power to indicate a root. For example, if the answer was $\\sqrt[3]{\\frac{35}{11}}$, then enter (35/11)^(1/3).

If $\\simplify{{cxcoeff}z^{cpower}+{cb}}=\\var{cc}$, then $z=$ [[0]], or [[1]].

Given $\\simplify{{cxcoeff}z^{cpower}+{cb}}=\\var{cc}$, we can rearrange the equation to get $z^\\var{cpower}$ by itself and then we can take the plus or minus $\\var{cpower}$nd rd th root to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{{cxcoeff}z^{cpower}+{cb}}$	$=$	$\\var{cc}$

$\\simplify{{cxcoeff}z^{cpower}}$	$=$	$\\simplify[basic]{{cc}-{cb}}$

$\\simplify{{cxcoeff}z^{cpower}}$	$=$	$\\simplify{{cc-cb}}$

$z^\\var{cpower}$	$=$	$\\simplify[!basic]{{cc-cb}/{cxcoeff}}$

\n $z^\\var{cpower}$ \n	\n $=$ \n	\n $\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}$ \n

$\\sqrt[\\var{cpower}]{z^\\var{cpower}}$	$=$	$\\pm\\sqrt[\\var{cpower}]{\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}}$

$z$	$=$	$\\pm\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}^{1/{cpower}}}$

That is, $z$ equals $-\\sqrt[\\var{cpower}]{\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}}$, or $\\sqrt[\\var{cpower}]{\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "({(cc-cb)/cxcoeff})^(1/{cpower})", "answerSimplification": "fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, 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check_pre_submit(gaps[gap_number][\"path\"], answer, exec_path)\n ),\n [gap_number,answer_number],\n zip(gap_order,answer_order)\n )", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

For this question input the exact value by using a fractional power to indicate a root. For example, if the answer was $\\sqrt[3]{\\frac{35}{11}}$, then enter (35/11)^(1/3).

If $\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}=\\var{dc}$, then $z=$ [[0]], or [[1]].

Given $\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}=\\var{dc}$, we can rearrange the equation to get $\\simplify{(z+{db})^{bpower}}$ by itself, then we can take the plus or minus $\\var{bpower}$nd rd th root of both sides, and then rearrange to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}$	$=$	$\\var{dc}$

$\\simplify{(z+{db})^{bpower}}$	$=$	$\\simplify[basic]{{dc}*{ddenom}}$

$\\simplify{(z+{db})^{bpower}}$	$=$	$\\var{dc*ddenom}$

$\\sqrt[\\var{bpower}]{\\simplify{(z+{db})^{bpower}}}$	$=$	$\\pm\\sqrt[\\var{bpower}]{\\var{dc*ddenom}}$

$\\simplify{z+{db}}$	$=$	$\\pm\\sqrt[\\var{bpower}]{\\var{dc*ddenom}}$

$z$	$=$	$\\pm\\simplify{root({dc*ddenom},{bpower})-{db}}$

$z$	$=$	$\\pm\\simplify{{dc*ddenom}^(1/{bpower})-{db}}$

That is, $z$ equals $-\\simplify{root({dc*ddenom},{bpower})-{db}}$, or $\\simplify{root({dc*ddenom},{bpower})-{db}}$.

Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers.

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By using the definition of the square root you should see that $(\\sqrt{q})^2=q$.

By using index laws you should see that $(q^{1/2})^2=q$.

The above equations imply that $\\sqrt{q}$ can also be written as [[0]].

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

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Given \\[(\\sqrt{q})^2=q=(q^{1/2})^2\\]

we can say \\[\\sqrt{q}=q^{1/2}\\]

Which we would type in as $q\\wedge(1/2)$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\n///var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// We just want to check that the student has written \"something to the power of something\"\n//var rule = Numbas.jme.compile('?? ^ ??');\n\n// Check the student's answer matches the pattern. \n//var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n\n// If not, take away marks\n//if(!m) {\n //this.multCredit(0,'Your answer is not in the form $x^y$.');\n//}\n", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "q^(1/2)", "answerSimplification": "basic", "showPreview": false, "checkingType": "absdiff", "checkingAccuracy": "0.00001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "maxlength": {"length": "11", "partialCredit": 0, "message": "

Your answer is longer than necessary.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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By using the definition of the cube root you should see that $(\\sqrt[3]{p})^3=p$.

By using index laws you should see that $(p^{1/3})^3=p$.

The above equations imply that $\\sqrt[3]{p}$ can also be written as [[0]].

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

Given \\[(\\sqrt[3]{p})^3=p=(p^{1/3})^3\\]

we can say \\[\\sqrt[3]{p}=p^{1/3}\\]

Which we would type in as $p\\wedge(1/3)$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\n///var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// We just want to check that the student has written \"something to the power of something\"\n//var rule = Numbas.jme.compile('?? ^ ??');\n\n// Check the student's answer matches the pattern. \n//var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n\n// If not, take away marks\n//if(!m) {\n //this.multCredit(0,'Your answer is not in the form $x^y$.');\n//}\n", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "p^(1/3)", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.00001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "maxlength": {"length": "11", "partialCredit": 0, "message": "

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

$\\sqrt[\\var{root1}]{g}$ = [[0]]

In general, we have $\\sqrt[n]{a}=a^{\\frac{1}{n}}$.

In particular, $\\sqrt[\\var{root1}]{g}=g^\\frac{1}{\\var{root1}}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\n///var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// We just want to check that the student has written \"something to the power of something\"\n//var rule = Numbas.jme.compile('?? ^ ??');\n\n// Check the student's answer matches the pattern. \n//var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n\n// If not, take away marks\n//if(!m) {\n //this.multCredit(0,'Your answer is not in the form $x^y$.');\n//}\n", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "g^(1/{root1})", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.00001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "maxlength": {"length": "11", "partialCredit": 0, "message": "

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

$\\displaystyle\\left(\\sqrt[\\var{root2}]{e}\\right)^\\var{power2}$ = [[0]]

Convert the root to a fractional power and then use the index laws to deal with the two different powers.

That is, \\begin{align}\\left(\\sqrt[\\var{root2}]{e}\\right)^\\var{power2}
&=\\left(e^{\\frac{1}{\\var{root2}}}\\right)^\\var{power2}\\\\
&=e^{\\frac{1}{\\var{root2}}\\times\\var{power2}}\\\\
&=e^{\\var[fractionnumbers]{power2/root2}}.\\end{align}

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\n///var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// We just want to check that the student has written \"something to the power of something\"\n//var rule = Numbas.jme.compile('?? ^ ??');\n\n// Check the student's answer matches the pattern. \n//var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n\n// If not, take away marks\n//if(!m) {\n //this.multCredit(0,'Your answer is not in the form $x^y$.');\n//}\n", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "e^({power2}/{root2})", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.00001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "maxlength": {"length": "11", "partialCredit": 0, "message": "

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

$\\sqrt[\\var{root3}]{r^\\var{power3}}$ = [[0]]

Convert the root to a fractional power and then use the index laws to deal with the two different powers.

That is, \\begin{align}\\left(\\sqrt[\\var{root3}]{r}\\right)^\\var{power3}
&=\\left(r^{\\frac{1}{\\var{root3}}}\\right)^\\var{power3}\\\\
&=r^{\\frac{1}{\\var{root3}}\\times\\var{power3}}\\\\
&=r^{\\var[fractionnumbers]{power3/root3}}.\\end{align}

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers.

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2..6

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$\\left(d^\\var{powers1}\\right)^2$ = [[0]]

We can use the index law $\\displaystyle(a^b)^c=a^{bc}$.

\\begin{align}\\left(d^\\var{powers1}\\right)^2&=d^{\\var{powers1}\\times 2}\\\\
&=d^\\var{2*powers1}\\end{align}

Alternatively:

\\begin{align}\\left(d^\\var{powers1}\\right)^2&=\\left(d^\\var{powers1}\\right)\\times\\left(d^\\var{powers1}\\right)\\\\
&=d^{\\var{powers1}+\\var{powers1}}\\\\&=d^\\var{2*powers1}\\end{align}

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\n///var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// We just want to check that the student has written \"something to the power of something\"\n//var rule = Numbas.jme.compile('?? ^ ??');\n\n// Check the student's answer matches the pattern. \n//var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n\n// If not, take away marks\n//if(!m) {\n //this.multCredit(0,'Your answer is not in the form $x^y$.');\n//}\n", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "d^{prodpow1}", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "maxlength": {"length": "4", "partialCredit": 0, "message": "

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$\\displaystyle(j^\\var{powers2[0][0]})^\\var{powers2[1][0]}$ = [[0]]

Interpret the powers in expanded form and then write the result in index form, or use the index law $\\displaystyle(a^b)^c=a^{bc}$.

\\begin{align}(j^\\var{powers2[0][0]})^\\var{powers2[1][0]}&=j^{\\var{powers2[0][0]}\\times\\var{powers2[1][0]}}\\\\
&=j^\\var{powers2[0][0]*powers2[1][0]}\\end{align}

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

$\\displaystyle\\left(\\left(m^\\var{powers3[0][0]}\\right)^\\var{powers3[1][0]}\\right)^\\var{powers3[2][0]}$ = [[0]]

Interpret the powers in expanded form and then write the result in index form, or use the index law $\\displaystyle(a^b)^c=a^{bc}$.

We could work from the inside to the outside:

\\begin{align}\\left(\\left(m^\\var{powers3[0][0]}\\right)^\\var{powers3[1][0]}\\right)^\\var{powers3[2][0]}&=\\left(m^{\\var{powers3[0][0]}\\times \\var{powers3[1][0]}}\\right)^\\var{powers3[2][0]}\\\\
&=\\left(m^{\\var{powers3[0][0]*powers3[1][0]}}\\right)^\\var{powers3[2][0]}\\\\
&=m^{\\var{powers3[0][0]*powers3[1][0]}\\times\\var{powers3[2][0]}}\\\\
&=m^{\\var{powers3[0][0]*powers3[1][0]*powers3[2][0]}}\\\\
\\end{align}

We could also work from the outside to the inside:

\\begin{align}\\left(\\left(m^\\var{powers3[0][0]}\\right)^\\var{powers3[1][0]}\\right)^\\var{powers3[2][0]}&=\\left(m^\\var{powers3[0][0]}\\right)^{\\var{powers3[1][0]}\\times\\var{powers3[2][0]}}\\\\
&=\\left(m^\\var{powers3[0][0]}\\right)^{\\var{powers3[1][0]*powers3[2][0]}}\\\\
&=m^{\\var{powers3[0][0]}\\times\\var{powers3[1][0]*powers3[2][0]}}\\\\
&=m^{\\var{powers3[0][0]*powers3[1][0]*powers3[2][0]}}\\\\
\\end{align}

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$\\displaystyle\\left(\\left(n^\\var{nint}\\right)^\\var{ndec}\\right)^{\\var{num}/\\var{den}}$ = [[0]]

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

Working from the outside to the inside, using the index law $\\displaystyle(a^b)^c=a^{bc}$:

\\begin{align}\\left(\\left(n^\\var{nint}\\right)^\\var{ndec}\\right)^{\\var{num}/\\var{den}}
&=\\left(n^\\var{nint}\\right)^{\\var{ndec}\\times\\frac{\\var{num}}{\\var{den}}}\\\\
&=\\left(n^\\var{nint}\\right)^{\\var[fractionnumbers]{ndec*num/den}}\\\\
&=n^{\\var{nint}\\times\\var[fractionnumbers]{ndec*num/den}}\\\\
&=n^{\\var[fractionnumbers]{nint*ndec*num/den}}
\\end{align}

Note the answer can be written using a fractional power or a decimal power but a fractional power is more commonly used.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers. Use negative powers if necessary.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"minpow4": {"name": "minpow4", "group": "Ungrouped variables", "definition": "min(powers3[0][0],powers3[2][0])", "description": "", "templateType": "anything", "can_override": false}, "neg": {"name": "neg", "group": "Ungrouped variables", "definition": "random(-12..-1)", "description": "", "templateType": "anything", "can_override": false}, "ndec": {"name": "ndec", "group": "Ungrouped variables", "definition": "random(-0.9..-0.1#0.1)", "description": "", "templateType": "anything", "can_override": false}, "diffpow3": {"name": "diffpow3", "group": "Ungrouped variables", "definition": "ndec-2*neg/2", "description": "", "templateType": "anything", "can_override": false}, "diffpow2": {"name": "diffpow2", "group": "Ungrouped variables", "definition": "powers2[0][0]+1-powers2[1][0]-powers2[2][0]", "description": "", "templateType": "anything", "can_override": false}, "diffpow1": {"name": "diffpow1", "group": "Ungrouped variables", "definition": "powers1[0][0]-powers1[1][0]", "description": "", "templateType": "anything", "can_override": false}, "diffpow4": {"name": "diffpow4", "group": "Ungrouped variables", "definition": "powers3[0][0]-powers3[2][0]", "description": "", "templateType": "anything", "can_override": false}, "powers1": {"name": "powers1", "group": "Ungrouped variables", "definition": "shuffle([[2,\"two\"],[3,\"three\"],[4,\"four\"],[5,\"five\"],[6,\"six\"]])[0..2]", "description": "

2..6

", "templateType": "anything", "can_override": false}, "powers3": {"name": "powers3", "group": "Ungrouped variables", "definition": "shuffle([[2,\"two\"],[3,\"three\"],[4,\"four\"],[5,\"five\"],[6,\"six\"]])[0..3]", "description": "", "templateType": "anything", "can_override": false}, "powers2": {"name": "powers2", "group": "Ungrouped variables", "definition": "shuffle([[2,\"two\"],[3,\"three\"],[4,\"four\"],[5,\"five\"],[6,\"six\"]])[0..3]", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["powers1", "diffpow1", "powers2", "diffpow2", "ndec", "neg", "diffpow3", "powers3", "diffpow4", "minpow4"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$u^\\var{powers1[0][0]}\\div u^\\var{powers1[1][0]}$ = [[0]]

Write the division as a fraction and cancel common factors or use the index law $\\displaystyle\\frac{a^b}{a^c}=a^{b-c}$.

Note, negative indices mean you divided by more than you had, for example, $\\displaystyle \\frac{1}{12^3}$ can be written as $12^{-3}$.

For our particular question we have, $u^\\var{powers1[0][0]}\\div u^\\var{powers1[1][0]}=\\frac{u^\\var{powers1[0][0]}}{u^\\var{powers1[1][0]}}$ is equal to $u^{\\var{powers1[0][0]}-\\var{powers1[1][0]}}=u^{\\var{diffpow1}}$.

Use ^ for powers. Input your answer in index form.

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$\\displaystyle\\frac{v^\\var{powers2[0][0]}\\times v}{v^\\var{powers2[1][0]} \\times v^\\var{powers2[2][0]}}$ = [[0]]

Cancel common factors or use the index laws, e.g. $\\displaystyle\\frac{a^b}{a^c}=a^{b-c}$ and $a^b\\times a^c = a^{b+c}$.

Note, negative indices mean you divided by more than you had, for example, $\\displaystyle \\frac{1}{12^3}$ can be written as $12^{-3}$.

For $\\frac{v^\\var{powers2[0][0]}\\times v}{v^\\var{powers2[1][0]} \\times v^\\var{powers2[2][0]}}$ we will simplify the numerator and denominator by using $a^b\\times a^c = a^{b+c}$ and end up with $\\frac{v^\\var{powers2[0][0]+1}}{v^\\var{powers2[1][0]+powers2[2][0]}}$. Then we will use $\\frac{a^b}{a^c}=a^{b-c}$ to finally get $v^\\var{powers2[0][0]+1-powers2[1][0]-powers2[2][0]}$.

Use the same approach you used in the above questions to simplify the following in index form.

$\\displaystyle\\frac{w^\\var{ndec}}{w^\\var{neg}}$ = [[0]]

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

Since the bases are all the same ($w$) and we are dividing, we can subtract the powers since in general, we have $\\displaystyle\\frac{a^b}{a^c}=a^{b-c}$. Also recall that subtracting a negative is equivalent to adding a positive.

$\\begin{align}\\frac{w^\\var{ndec}}{w^\\var{neg}}&=w^{\\var{ndec}-\\var{neg}}\\\\
&=w^{\\var{ndec}+\\var{abs(neg)}}\\\\&=w^{\\var{ndec+abs(neg)}}\\end{align}$

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$x^\\var{powers3[0][0]}\\times y^\\var{powers3[1][0]} \\div x^\\var{powers3[2][0]}$ = [[0]]

It is important to note that the bases are different! Index laws can only be applied if the bases are the same (or can be made the same). Because of this we deal with the different bases separately.

Notice the first part of the expression can not be simplified using index laws.

$x^\\var{powers3[0][0]}\\times y^\\var{powers3[1][0]}$

However, with the division we can do some simplification. We can either:

write it as a fraction and cancel the common factor of $x^\\var{minpow4}$ from the top and bottom:
\\[\\frac{x^\\var{powers3[0][0]}\\times y^\\var{powers3[1][0]}}{ x^\\var{powers3[2][0]}}=\\frac{x^\\var{powers3[0][0]-minpow4}\\times y^\\var{powers3[1][0]}}{ x^\\var{powers3[2][0]-minpow4}}=x^\\var{diffpow4}y^\\var{powers3[1][0]}\\]
Subtract the powers, \"top power minus the bottom power\" for the terms with the same base:
\\[\\frac{x^\\var{powers3[0][0]}\\times y^\\var{powers3[1][0]}}{ x^\\var{powers3[2][0]}}=x^\\var{diffpow4}y^\\var{powers3[1][0]}\\]

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\n///var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// We just want to check that the student has written \"something to the power of something\"\n//var rule = Numbas.jme.compile('?? ^ ??');\n\n// Check the student's answer matches the pattern. \n//var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n\n// If not, take away marks\n//if(!m) {\n //this.multCredit(0,'Your answer is not in the form $x^y$.');\n//}\n", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "x^{diffpow4}*y^{powers3[1][0]}", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.00001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": true, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "musthave": {"strings": ["^"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Is the following statement true or false?

$\\displaystyle\\frac{2z}{z^\\var{powers1[0][0]}} = 2^\\var{-powers1[0][0]}$

Index laws only can be applied if the bases are the same (or can be made the same). We can apply the index law to the left-hand side but the right-hand side has a different base so it is not the correct expression.

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Is the following statement true or false?

$\\displaystyle\\frac{2a}{a^\\var{powers1[0][0]}} = 2^\\var{1-powers1[0][0]}$

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False

Is the following statement true or false?

$\\displaystyle\\frac{2b}{b^\\var{powers1[0][0]}} = 2\\times b^\\var{1-powers1[0][0]}$

Since the bases are all the same ($b$) and we are dividing, we can simply subtract the powers.

\\[\\frac{2b}{b^\\var{powers1[0][0]}}=\\frac{2\\times b^1}{b^\\var{powers1[0][0]}} = 2\\times b^{1-\\var{powers1[0][0]}}= 2 b^\\var{1-powers1[0][0]}\\]

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Use ^ to signify powers.

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2..6

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For an insight into negative indices, consider the following table:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

index form	$b^3$	$b^2$	$b^1$	$b^0$	$b^{-1}$	$b^{-2}$
meaning	$b\\times b \\times b$	$b \\times b$	$b$	$1$	[[0]]	[[1]]

Notice each time the power decreases by $1$, the result is divided by $b$. Using this idea, and / for division, fill in the rest of the table.

Each time you reduce the power by 1 you divide the result by the base, that is $b$. Following this pattern:

\\[b^{-1}=\\frac{1}{b}\\]

and

\\[b^{-2}=\\frac{1}{b}\\div b=\\frac{1}{b^2}\\]

Note you could input the second expression as $1/b/b$ or $1/b\\wedge2$ or $1/b^2$.

Don't use negative powers here in this table.

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Don't use negative powers here in this table.

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The fraction $\\displaystyle\\frac{1}{n^\\var{power1}}$ can be written using a negative index as [[0]].

Notice:
\\[1=n^0=n^{\\var{power1}-\\var{power1}}=n^{\\var{power1}} n^{-\\var{power1}}\\]

That is, we have

\\[1=n^{\\var{power1}} n^{-\\var{power1}}\\]

by dividing both sides of this equation by $n^{\\var{power1}}$ we get

\\[\\frac{1}{n^{\\var{power1}}}=n^{-\\var{power1}}\\]

In general, we have $\\displaystyle\\frac{1}{a^c}=a^{-c}$.

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Use ^ for powers. Input your answer in index form.

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Write your answer with a negative index.

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The expression $h^{\\var{power2}}$ can be written without a negative index as the fraction [[0]].

Notice:
\\[1=h^0=h^{\\var{power2}+\\var{-power2}}=h^{\\var{power2}} h^{\\var{-power2}}\\]

That is, we have

\\[1=h^{\\var{power2}} h^{\\var{-power2}}\\]

by dividing both sides of this equation by $h^{\\var{-power2}}$ we get

\\[\\frac{1}{h^{\\var{-power2}}}=h^{\\var{power2}}\\]

In general, we have $\\displaystyle\\frac{1}{a^c}=a^{-c}$.

Use ^ for powers. Input your answer in index form.

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The expression $\\displaystyle\\frac{1}{k^{\\var{-power3}}}$ can be written without a negative index and without the use of a fraction.

The simplest way to write $\\displaystyle\\frac{1}{k^{\\var{-power3}}}$ in index form would be [[0]]

You can think of the negative power as forcing the term to the other part of the fraction (if it was on top it goes to the bottom and if it was on the bottom it goes to the top) except now it has a positive power.

We can see this by using our rules for dividing fractions:

\\[\\frac{1}{k^\\var{-power3}}=\\frac{1}{\\left(\\frac{1}{k^\\var{power3}}\\right)}=1\\div{\\left(\\frac{1}{k^\\var{power3}}\\right)}=1\\times \\left(\\frac{{k^\\var{power3}}}{1}\\right)=k^\\var{power3}\\]

In general, we have $\\displaystyle\\frac{1}{a^c}=a^{-c}$ and $\\displaystyle\\frac{1}{a^{-c}}=a^{c}$.

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Use index form.

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Don't use fractions or negative indices.

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers.

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2..6

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$ w ^\\var{powers1[0][0]}\\times w ^\\var{powers1[1][0]} \\times w ^\\var{powers1[2][0]}$ = [[0]]

Recall:

$w^\\var{powers1[0][0]}$ means $\\var{powers1[0][1]}$ $w$s all multiplied together.
$w^\\var{powers1[1][0]}$ means $\\var{powers1[1][1]}$ $w$s all multiplied together.
$w^\\var{powers1[2][0]}$ means $\\var{powers1[2][1]}$ $w$s all multiplied together.

So in total how many $w$s are there multiplied together?

Well, $\\var{powers1[0][0]}+\\var{powers1[1][0]}+\\var{powers1[2][0]}=\\var{sumpow1}$. And so our answer is $w^\\var{sumpow1}$.

Note, in general $a^ba^c=a^{b+c}$.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$x^\\var{powers2[0][0]}\\times x^\\var{powers2[1][0]} \\times x^\\var{powers2[2][0]}$ = [[0]]

Recall:

$x^0$ means no $x$s all multiplied together. (As you know, $x^0=1$).
$x^1$ means just one $x$. (As you know, $x^1=x$)
$x^\\var{wild[0]}$ means $\\var{wild[1]}$ $x$s all multiplied together.

So in total how many $x$s are there multiplied together?

Well, $0+1+\\var{wild[0]}=\\var{sumpow2}$. And so our answer is $x^\\var{sumpow2}$.

Note, in general $a^ba^c=a^{b+c}$, $a^0=1$ and $a^1=a$.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

$\\displaystyle a^\\var{dec}\\times a^\\var{neg} \\times a^{\\var{num}/\\var{den}}$ = [[0]]

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

Since the bases are all the same ($a$) and we are multiplying, we can simply add the powers.

You could convert the fraction to a decimal and then add them all. Or you could add them all as fractions.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$z^\\var{powers3[0][0]}\\times y^\\var{wild[0]} \\times z^\\var{powers3[1][0]}$ = [[0]]

It is important to note that the bases are different! We can only add the powers if the bases are the same.

Recall:

$z^\\var{powers3[0][0]}$ means $\\var{powers3[0][1]}$ $z$s all multiplied together.
$z^\\var{powers3[1][0]}$ means $\\var{powers3[1][1]}$ $z$s all multiplied together.
$y^\\var{wild[0]}$ means $\\var{wild[1]}$ $y$s all multiplied together.

So in total what do we have?

$y^\\var{wild[0]}z^\\var{sumpow4}$.

Note we would type y^{wild[0]}*z^{sumpow4}.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\n///var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// We just want to check that the student has written \"something to the power of something\"\n//var rule = Numbas.jme.compile('?? ^ ??');\n\n// Check the student's answer matches the pattern. \n//var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n\n// If not, take away marks\n//if(!m) {\n //this.multCredit(0,'Your answer is not in the form $x^y$.');\n//}\n", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "y^{wild[0]}*z^{sumpow4}", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": true, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "musthave": {"strings": ["^"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Is the following statement true or false?

$m\\times p^\\var{powers1[0][0]} = (mp)^\\var{powers1[0][0]}$

It is important to note that the bases are different! Index laws only can be applied if the bases are the same (or can be made the same).

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True

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False

Is the following statement true or false?

$q\\times u^\\var{powers1[0][0]} = (qu)^\\var{powers1[0][0]+1}$

It is important to note that the bases are different! Index laws only can be applied if the bases are the same (or can be made the same). We can only add the powers if the bases are the same.

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False

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Use ^ to signify indices.

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The expression $\\var{latex(expanded[0])}$ is in expanded form, the same expression in index form is [[0]].

The expression $\\var{latex(expanded[0])}$ written in index form is $a^\\var{expanded[1]}$.

The $a$ is the base and $\\var{expanded[1]}$ is the power/exponent/index. The power signifies how many bases are multiplied together.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\n//var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// We just want to check that the student has written \"something to the power of something\"\n//var rule = Numbas.jme.compile('?? ^ ??');\n\n// Check the student's answer matches the pattern. \n//var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n\n// If not, take away marks\n//if(!m) {\n//this.multCredit(0,'Your answer is not in the form $x^y$.');\n//}\n", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "a^{expanded[1]}", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "musthave": {"strings": ["^"], "showStrings": false, "partialCredit": 0, "message": "

You need to use indices. Use ^ to signify indices.

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Nice try!

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The expression $b^1$ is normally written as [[0]]

$b^1$ just means there is one $b$. So we normally don't write the power. We normally just write $b$.

True or False:

The expression $\\displaystyle\\left(\\frac{c}{x}\\right)^\\var{p2}$ is equivalent to $\\displaystyle\\frac{c^\\var{p2}}{x}$.

The power is acting on the whole fraction, for example:

\\[\\left(\\frac{c}{x}\\right)^2=\\left(\\frac{c}{x}\\right)\\times\\left(\\frac{c}{x}\\right)=\\frac{c^2}{x^2}\\]

In general $\\left(\\frac{a}{b}\\right)^n=\\frac{a^n}{b^n}$

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False

True or False:

The expression $\\displaystyle\\left(m\\times n\\right)^\\var{p1}$ is equivalent to $\\displaystyle m^\\var{p1} n^\\var{p1}$.

The power is acting on the whole bracket, for example:

\\[\\left(m\\times n\\right)^2=\\left(m\\times n\\right)\\times\\left(m\\times n\\right)=m^2\\times n^2\\]

In general $\\left(ab\\right)^n=a^n b^n$

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We start solving the equation one operation at a time by doing the inverse to both sides, when we get to undoing the exponential we apply a log to both sides, we then use a log law and continue solving.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a}$	$=$	$\\simplify{{p}({b})^(n/{d})+{c}}$
$\\simplify{{a-c}}$	$=$	$\\simplify{{p}({b})^(n/{d})}$	(subtract $\\var{c}$ from both sides)
$\\var{frac}$	$=$	$\\simplify{{b}^(n/{d})}$	(divide both sides by $\\var{p}$)
$\\log(\\var{frac})$	$=$	$\\log(\\var{b}^{\\frac{n}{\\var{d}}})$	(take the log of both sides)
	$=$	$\\frac{n}{\\var{d}}\\log(\\var{b})$	(use a log law)

$\\displaystyle{\\frac{\\log(\\var{frac})}{\\log(\\var{b})}}$	$=$	$\\frac{n}{\\var{d}}$	(divide both sides by $\\log(\\var{b})$)

$\\displaystyle{\\frac{\\var{d}\\log(\\var{frac})}{\\log(\\var{b})}}$	$=$	$n$	(multiply both sides by $\\var{d}$)

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

${\\var{FV}}$	$=$	$\\displaystyle{\\simplify{{pay}((1+{int})^n-1)/{int}}}$
$\\simplify{{FV*int}}$	$=$	$\\displaystyle{\\simplify{{pay}((1+{int})^n-1)}}$	(multiply both sides by $\\var{int}$)
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int}/{pay}}}$	$=$	$\\displaystyle{\\simplify{(1+{int})^n-1}}$	(divide both sides by $\\var{pay}$)
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int}/{pay}+1}}$	$=$	$\\displaystyle{\\simplify{(1+{int})^n}}$	(add $1$ to both sides)
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}}$	$=$	$\\displaystyle{\\simplify{(1+{int})^n}}$	(tidy up left hand side)

$\\displaystyle{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}$	$=$	$\\displaystyle{\\log\\left((\\var{1+int})^n\\right)}$	(take the log of both sides)

$\\displaystyle{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}$	$=$	$\\displaystyle{n\\log(\\var{1+int})}$	(use a log law)

$\\displaystyle{\\frac{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}{\\log(\\var{1+int})}}$	$=$	$n$	(divide both sides by $\\log(\\var{1+int})$)

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Solve the following equation for $n$

$\\begin{align*}\\simplify{{a}={p}({b})^(n/{d})+{c}}.\\end{align*}$

$n=$ [[0]]

Note: Typing $\\log(5)$ will input the value $\\log_{10}(5)$, whereas $\\log5$ will not work.
Note: Typing $\\ln(5)$ will input the value $\\log_e(5)$, whereas $\\ln5$ will not work.

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Solve the following equation for $n$

$\\displaystyle{\\simplify{{FV}={pay}((1+{int})^n-1)/{int}}}.$

$n=$ [[0]]

Note: Typing $\\log(5)$ will input the value $\\log_{10}(5)$, whereas $\\log5$ will not work.
Note: Typing $\\ln(5)$ will input the value $\\log_e(5)$, whereas $\\ln5$ will not work.

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A very simple algebraic fraction multiplied by a whole number. No cancelling is required by design.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express the following as a single fraction. Use / as the fraction bar, use brackets to group the denominator and use * for multiplication between a term and a bracket, e.g. $\\dfrac{7(m+1)}{2n}$ is written 7*(m+1)/(2n)

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$\\displaystyle \\simplify[alwaysTimes,!simplifyFractions]{(w+{f})/{j} * {d}}=$[[0]]

We can write the whole number as a fraction over $1$, then we can multiply the numerators and multiply the denominators. We can cancel any common factors before or after multiplication.

$\\begin{align*}\\displaystyle \\simplify[alwaysTimes,!simplifyFractions]{(w+{f})/{j} * {d}}&=\\simplify[alwaysTimes,!simplifyFractions]{(w+{f})/{j} * ({d}/1)}\\\\[3pt]&=\\simplify[alwaysTimes,!simplifyFractions]{(w+{f})*{d}/({j} * 1)}\\\\[3pt]&=\\simplify{({d}w+{d*f})/{j}}\\end{align*}$

Note, there are no common factors to cancel.

A whole number divided by a very simple algebraic fraction. No cancelling is required by design.

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Evaluate the following and write your answer as a single fraction. Use / to signify a fraction or division, for example $\\frac{2a-1}{x+3}$ is written (2a-1)/(x+3). Simplify/cancel where possible.

determines how the division is displayed

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$\\displaystyle \\simplify[alwaysTimes,!simplifyFractions]{{d}/((x+{f})/{j})}=\\;$$\\displaystyle \\var{d} \\div \\simplify{((x+{f})/{j})}=\\;$[[0]]

Note that by looking at the length of the fraction bars we can determine that $\\displaystyle \\simplify[alwaysTimes,!simplifyFractions]{{d}/((x+{f})/{j})}$ represents $\\displaystyle \\var{d} \\div \\simplify{(x+{f})/{j}}$.

We can write the whole number as a fraction over $1$, and then we can do the division by multiplying by the reciprocal. We can cancel any common factors before or after multiplication.

$\\begin{align*}\\var{d} \\div \\simplify{(x+{f})/{j}}&=\\frac{\\var{d}}{1}\\div\\simplify{(x+{f})/{j}}\\\\[3pt]&=\\frac{\\var{d}}{1}\\times\\simplify{{j}/(x+{f})}\\\\[3pt]&=\\simplify[alwaysTimes,!simplifyFractions]{{d}*{j}/(1*(x+{f}))}\\\\[3pt]&=\\simplify{{d*j}/(x+{f})}\\end{align*}$

Note, there are no common factors to cancel.

Fractions don't have a common denominator. Need to find one. Addition and subtraction 50:50 split.

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Express the following as a single fraction. Use / as the fraction bar and use brackets to group the numerator and denominator separately, e.g. $\\dfrac{m+1}{2n}$ is written (m+1)/(2n)

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adding or subtracting

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common factor of denominators

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extra factors in the denominators

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$\\displaystyle\\simplify{({a}x+{b})/({cf*ee}x)+{addsub}*(({c}x+{d}y)/({cf*f}x*y))}=$[[0]]

We need to get the denominators to be the same, preferably the lowest common denominator, at which point, we can simply addsubtract the new numerators and put the result over the common denominator.

Since the first denominator equals $\\var{cf}\\times\\var{ee}\\times x$ and the second equals $\\var{cf}\\times \\var{f}\\times x\\times y$, the lowest common denominator will be the product $\\var{cf}\\times \\var{ee}\\times \\var{f}\\times x\\times y$, that is, $\\var{ansden}xy$. Therefore, we will multiply the first fraction by $\\frac{\\var{f}y}{\\var{f}y}$ and the second fraction by $\\frac{\\var{ee}}{\\var{ee}}$ as follows.

Since the first denominator equals $\\var{ee}\\times x$ and the second equals $\\var{f}\\times x\\times y$, the lowest common denominator will be the product $\\var{ee}\\times \\var{f}\\times x\\times y$, that is, $\\var{ansden}xy$. Therefore, we will multiply the first fraction by $\\frac{\\var{f}y}{\\var{f}y}$ and the second fraction by $\\frac{\\var{ee}}{\\var{ee}}$ as follows.

$\\begin{align*}\\displaystyle\\simplify{({a}x+{b})/({cf*ee}x)+{addsub}*(({c}x+{d}y)/({cf*f}x*y))}&=\\simplify{({a}x+{b})/({cf*ee}x)}\\times\\dfrac{\\var{f}y}{\\var{f}y}+\\simplify{(({c}x+{d}y)/({cf*f}x*y))}\\times\\dfrac{\\var{ee}}{\\var{ee}}\\\\[3pt]&=\\dfrac{\\simplify{{a*f}x*y+{b*f}y}}{\\var{ansden}xy}+\\dfrac{\\simplify{{c*ee}x+{d*ee}y}}{\\var{ansden}xy}\\\\[3pt]&=\\dfrac{\\simplify{{a*f}x*y+{b*f+addsub*ee*d}y+{addsub*c*ee}x}}{\\var{ansden}xy}\\end{align*}$

$\\begin{align*}\\displaystyle\\simplify{({a}x+{b})/({cf*ee}x)+{addsub}*(({c}x+{d}y)/({cf*f}x*y))}&=\\simplify{({a}x+{b})/({cf*ee}x)}\\times\\dfrac{\\var{f}y}{\\var{f}y}-\\simplify{(({c}x+{d}y)/({cf*f}x*y))}\\times\\dfrac{\\var{ee}}{\\var{ee}}\\\\[3pt]&=\\dfrac{\\simplify{{a*f}x*y+{b*f}y}}{\\var{ansden}xy}-\\dfrac{\\simplify{{c*ee}x+{d*ee}y}}{\\var{ansden}xy}\\\\[3pt]&=\\dfrac{\\simplify{{a*f}x*y+{b*f+addsub*ee*d}y+{addsub*c*ee}x}}{\\var{ansden}xy}\\end{align*}$

Things like \"expand 4(5a-3)\"

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The expression $\\simplify{{nmult}({nxcoeff}a+{nconstant})}$ is factorised (written as a product). We can expand the expression (so it is written as a sum) to get

[[0]]$a$ + [[1]]

The number in front of the bracket is multiplying the bracketed term, that is, each term in the brackets. Also, recall that a negative multiplied by a negative is a positive.

$\\begin{align*}
\\simplify{{nmult}({nxcoeff}a+{nconstant})}&=\\simplify[!noleadingminus]{{nmult}*{nxcoeff}a+{nmult} * {nconstant}}\\\\&=\\simplify[!noLeadingMinus]{{nmult*nxcoeff}a+{nmult*nconstant}}
\\end{align*}$

Things like \"expand -(2x+3y+4)\"

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Expand $ -(\\simplify[!noLeadingMinus]{{cx}x+{cy}y+{cc}})$.

[[0]] $x$ + [[1]] $y$ + [[2]]

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A negative sign in front of a bracket is a common way to signify $-1$ times the bracketed term. The result is that it changes the sign of each term in the brackets.

$-(\\var{cx}x-\\var{-cy}y+\\var{cc})=\\simplify[!noLeadingMinus]{{-cx}x+{-cy}y+{-cc}}$

Basically expand an expression like \"5y(-2z+3)\" where the student types answer into a single gap.

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Expanding $\\simplify{{a}*{expression(letters[0])}}\\left(\\simplify[!noleadingminus]{{b}*{expression(letters[1])}+{c}}\\right)$ gives [[0]].

The expression in front of the bracket is multiplying the bracketed term, that is, each term in the brackets.

That is

$\\begin{align*}
\\simplify{{a}*{expression(letters[0])}}\\left(\\simplify[!noleadingminus]{{b}*{expression(letters[1])}+{c}}\\right)
&=\\simplify[!collectNumbers, !constantsFirst, !basic]{({a}*{expression(letters[0])})*({b}*{expression(letters[1])})+({a}*{expression(letters[0])})*{c}}\\\\
&=\\simplify{{a}*{expression(letters[0])}*{b}*{expression(letters[1])}+{a}*{expression(letters[0])}*{c}}
\\end{align*}$

significant figures of velocity

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Given $a=\\var{aval}$, $b=\\var{bval}$ and $c=\\var{cval}$, the value of $b^2-4ac$ is If $x=\\var{xval}$ and $y=\\var{yval}$ then $\\simplify{(x-{aval})^2+(y-{cval})^2}$ = Suppose $m_0=\\var{m0}$, $v=\\var{v_sig_figs}\\times 10^8$, $c=3\\times 10^8$ and $m=\\dfrac{m_0}{\\sqrt{1-\\frac{v^2}{c^2}}}$. Then we have $m=$ [[0]] (rounded to two decimal places)

Be mindful of the order of operations and negatives when evaluating expressions.

Substituting $a=\\var{aval}$, $b=\\var{bval}$ and $c=\\var{cval}$ into $b^2-4ac$ gives

\\[(\\var{bval})^2-4(\\var{aval})(\\var{cval})=\\simplify[basic]{{bval^2}-4{aval}{cval}}=\\simplify[basic]{{bval^2}+{-4*aval*cval}}=\\var{disans}.\\]

Substituting $x=\\var{xval}$ and $y=\\var{yval}$ into $\\simplify{(x-{aval})^2+(y-{cval})^2}$ gives

\\[\\simplify[basic]{({xval}-{aval})^2+({yval}-{cval})^2}=(\\var{xval-aval})^2+(\\var{yval-cval})^2=\\var{(xval-aval)^2}+\\var{(yval-cval)^2}=\\var{cirans}.\\]

Substituting $m_0=\\var{m0}$, $v=\\var{v_sig_figs}\\times 10^8$ and $c=3\\times 10^8$ into $m=\\dfrac{m_0}{\\sqrt{1-\\frac{v^2}{c^2}}}$ gives

\\[m=\\dfrac{\\var{m0}}{\\sqrt{1-\\frac{(\\var{v_sig_figs}\\times 10^8)^2}{(3 \\times 10^8)^2}}}=\\var{mans} \\quad\\text{ (to two decimal places).}\\]

Students seem to freak out when their answer is not written exactly the same as the answer provided. This question tries to enforce that $(x-y)=-(y-x)$ and $\\frac{a-b}{c-d}=\\frac{b-a}{d-c}$

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This question asks you to compared different looking answers, and determine if they are equivalent.

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Suppose you do a maths question and your answer is

\\[y=\\var{a*b}+\\frac{\\var{c}-x}{\\var{d}}.\\]

However, your friend has an answer of

\\[y=\\var{a*b} \\var{sym1}\\frac{x-\\var{c}}{\\var{d}}.\\]

These answers are...

Consider doing the subtraction $11-25$. Often people do the easier subtraction $25-11$, get $14$, and then they put a negative in front of it to conclude $11-25=-14$. This works because

\\[11-25=-(25-11)=-14.\\]

So if we swap the order of subtraction, we need to put a negative out the front, but this is the same as just multiplying by $-1$ since $-(25-11)=-1\\times(25-11)$, which is also the same as dividing by $-1$.

Therefore, reversing the order of a subtraction is the same as multiplying (or dividing) by $-1$.

To determine if $\\frac{\\var{c}-x}{\\var{d}}$ is equal to $\\frac{x-\\var{c}}{\\var{d}}$, notice the only difference is the subtraction in the numerator is reversed. But $\\var{c}-x\\ne x-\\var{c}$. So these answers are not the same!

To determine if $\\frac{\\var{c}-x}{\\var{d}}$ is equal to $-\\frac{x-\\var{c}}{\\var{d}}$, notice

$\\begin{align}-\\frac{x-\\var{c}}{\\var{d}}&=\\frac{-(x-\\var{c})}{\\var{d}}\\\\&=\\frac{-x+\\var{c}}{\\var{d}}\\\\&=\\frac{\\var{c}-x}{\\var{d}}\\end{align}$

So the negative out the front and the reversing of the subtraction cancelled each other out, and these answers are actually the same.

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equal!

", "

not equal!

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Suppose you do a maths question and your answer is

\\[z=\\frac{\\var{a}x^\\var{p}-\\var{b}y^\\var{q}}{\\var{c}xy-\\var{d}x^\\var{r}y^\\var{s}}.\\]

However, your friend has an answer of

\\[z=\\frac{\\var{b}y^\\var{q}-\\var{a}x^\\var{p}}{\\var{d}x^\\var{r}y^\\var{s}\\var{sym2}\\var{c}xy}.\\]

These fractions are...

[[0]]

You should notice that these fractions are very similar except that the order of subtraction is reversed in the numerator and the denominator. We should know that reversing the order of subtraction introduces a negative out the front, if we do this twice we will have two negatives out the front, which of course means a positive! That is,

$\\begin{align}\\frac{\\var{a}x^\\var{p}-\\var{b}y^\\var{q}}{\\var{c}xy-\\var{d}x^\\var{r}y^\\var{s}}&=\\frac{-(\\var{b}y^\\var{q}-\\var{a}x^\\var{p})}{\\var{c}xy-\\var{d}x^\\var{r}y^\\var{s}}\\\\&=\\frac{-(\\var{b}y^\\var{q}-\\var{a}x^\\var{p})}{-(\\var{d}x^\\var{r}y^\\var{s}-\\var{c}xy)}\\\\&=\\frac{\\var{b}y^\\var{q}-\\var{a}x^\\var{p}}{\\var{d}x^\\var{r}y^\\var{s}-\\var{c}xy}\\end{align}$

So the answers are the same!

You should notice that in the numerator the order of subtraction has been swapped and in the denominator a $-\\var{d}x^\\var{r}y^\\var{s}$ has been replaced with $+\\var{d}x^\\var{r}y^\\var{s}$. These are not the same answers. If you require further proof, set them to be equal and see what happens, or even easier, substitute a value for $x$ and $y$ into both of them:

Let $x=1$ and $y=1$ and we will compare the fractions. For 'your' answer we get $\\simplify[fractionNumbers,simplifyFractions]{{(a-b)/(c-d)}}$ but for 'your friends' answer we get $\\simplify[fractionNumbers,simplifyFractions]{{(a-b)/(c+d)}}$ and therefore the fractions are not equal!

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equivalent! Just multiply (or divide) the numerator and denominator by $-1$ to see this.

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Evaluate the following and write your answer as a single fraction. Use / to signify a fraction or division, for example $\\frac{2a-1}{x+3}$ is written (2a-1)/(x+3). Simplify/cancel where possible.

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$\\displaystyle\\frac{\\var{a}x}{\\var{b}}+\\frac{x+\\var{c}}{\\var{b}}=$ [[0]]

$\\displaystyle\\frac{\\var{d}}{\\var{c}y}-\\frac{\\var{a}}{\\var{c}y}=$ [[1]]

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Add the tops, leave the bottom the same.

These fractions have a common denominator (the number on the bottom). This means they are out of the same number of parts and can be compared easily, for example, it is clear $\\frac{2}{3}$ is less than $\\frac{5}{3}$ but not so clear that $\\frac{3}{5}$ is less than $\\frac{2}{3}$.

Let's say you need to evaluate $\\frac{2}{3}+\\frac{5}{3}$, in words this is 'two thirds plus five thirds', so how many thirds are there in total? Seven thirds!

So we have

\\[\\frac{2}{3}+\\frac{5}{3}=\\frac{2+5}{3}=\\frac{7}{3}\\]

The same logic is used for subtraction. Suppose you had seven fourths and someone borrowed three fourths, then you are left with four fourths.

That is

\\[\\frac{7}{4}-\\frac{3}{4}=\\frac{7-3}{4}=\\frac{4}{4}=1\\]

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$\\displaystyle\\simplify{(a+{f})/{g}+({h}a+1)/{j}}=$ [[0]]

$\\displaystyle\\simplify{(b+{h})/{f}-(b+{j})/{g}}=$ [[1]]

$\\displaystyle \\frac{\\var{a}}{\\var{d}r}+\\var{f}r=$ [[2]]

Rewrite the fractions so they have a common denominator. Then perform the addition or subtraction as required.

If your question was $\\frac{5}{4}+\\frac{3}{8}$ we could rewrite the first fraction as $\\frac{10}{8}$ (by multiplying the top and bottom by 2) and then both fractions would have a denominator of 8. At this point, we can perform the addition. Our working might look like this:

\\[\\frac{5}{4}+\\frac{3}{8}=\\frac{5\\times 2}{4\\times 2}+\\frac{3}{8}=\\frac{10}{8}+\\frac{3}{8}=\\frac{13}{8}\\]

Often we need to rewrite both fractions to get a common denominator, for instance, $\\frac{5}{4}-\\frac{2}{3}$. We could multiply the first fraction by 3 on the top and bottom, so that it's denominator was 12, and then multiply the second fraction by 4 on the top and bottom so that it also had a denominator of 12. Then we could perform the subtraction. Our working might look like this:

\\[\\frac{5}{4}-\\frac{2}{3}=\\frac{5\\times 3}{4\\times 3}-\\frac{2\\times 4}{3\\times 4}=\\frac{15}{12}-\\frac{8}{12}=\\frac{7}{12}\\]

Also, recall that whole numbers are just fractions with a denominator of 1, for example $3=\\frac{3}{1}$.

In general, the best denominator is the lowest common multiple (LCM) of the two denominators.

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$\\displaystyle\\frac{m+1}{n+1}\\times \\frac{y}{x}=$ [[0]]

$\\displaystyle -\\frac{\\var{f}+w}{\\var{j}}\\times \\var{d}=$ [[1]]

Multiply the tops and the bottoms.

For example

\\[\\frac{4}{5}\\times \\frac{2}{3}=\\frac{4\\times 2}{5 \\times 3}=\\frac{8}{15}\\]

Also recall that whole numbers are just fractions with a denominator of 1, for example $7=\\frac{7}{1}$.

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$\\displaystyle{\\simplify{({f}+{a}x)^2/{h}}}\\div \\simplify{(({f}+{a}x){g})/({j}x)}=$ [[0]]

$\\displaystyle \\frac{\\var{b}q}{\\var{c}q}\\div (\\var{d}+t)=$ [[1]]

$\\displaystyle \\var{j}z\\div \\left(\\frac{\\var{-d}(z+1)^2}{\\var{f}z}\\right)=$ [[2]]

Flip the second fraction and then multiply.

Flipping a fraction is also known as taking the reciprocal of the fraction (or inverting a fraction). Note that a whole number is also a fraction with a denominator of 1, for example, $6=\\frac{6}{1}$.

How do you find half of a number? You could 'divide it by 2', or you could 'multiply by $\\frac{1}{2}$. Notice that $\\frac{1}{2}$ is the reciprocal of 2. When we divide by a number this is actually the same as multiplying by its reciprocal.

Suppose you need to evaluate $\\frac{3}{7}\\div\\frac{5}{4}$. Recall this is the same as asking 'how many $\\frac{5}{4}$s are in $\\frac{3}{7}$?', but that doesn't seem to be very helpful here! What is helpful is realising that dividing by $\\frac{5}{4}$ is the same as multiplying by $\\frac{4}{5}$. Our working could look like this

\\[\\frac{3}{7}\\div\\frac{5}{4}=\\frac{3}{7}\\times\\frac{4}{5}=\\frac{3\\times 4}{7\\times 5}=\\frac{12}{35}\\]

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$\\displaystyle \\frac{\\frac{\\var{b}+x}{y+\\var{a}}}{\\frac{ \\var{b}}{\\var{a}}}=$ [[0]]

$\\displaystyle \\frac{\\frac{w+\\var{f}}{\\var{g}w}}{w+\\var{f}}=$ [[1]]

$\\displaystyle \\frac{\\var{j}r}{\\frac{\\var{h}r}{\\var{c}r}}=$ [[2]]

The fraction bar means division.

The fraction $\\frac{2}{3}$ means 2 divided by 3. So these questions are just division questions! It is important to note which fraction bar is big and which are small, so you know the order of the divisions.

Here are some examples:

\\[\\frac{7}{\\frac{5}{6}}=7\\div\\frac{5}{6} =7\\times\\frac{6}{5}=\\frac{42}{5}\\]

\\[\\frac{\\frac{7}{5}}{6}=\\frac{7}{5}\\div 6=\\frac{7}{5}\\times \\frac{1}{6}=\\frac{7}{30}\\]

\\[\\frac{\\frac{9}{11}}{\\frac{5}{3}}=\\frac{9}{11}\\div\\frac{5}{3}=\\frac{9}{11}\\times \\frac{3}{5}=\\frac{27}{55}\\]

Add, subtract, multiply and divide algebraic fractions.

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Learn from your mistakes and have another attempt by clicking on 'Try another question like this one' until you get full marks.

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Questions to test if the student knows the inverse of fractional power or root (and how to solve equations that contain them).

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Please complete the following.

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((bc-bb)/bxcoeff)^(1/bpower)

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intsoln^intpower

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If $\\sqrt[\\var{intpower}]{x}=\\var{intrhs}$, then $x=$ [[0]].

Given $\\sqrt[\\var{intpower}]{x}=\\var{intrhs}$, we raise both sides to the power of $\\var{intpower}$ to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\sqrt[\\var{intpower}]{x}$	$=$	$\\var{intrhs}$

$\\left(\\sqrt[\\var{intpower}]{x}\\right)^{\\var{intpower}}$	$=$	$\\simplify[basic]{({intrhs})^{intpower}}$

$x$	$=$	$\\var{intsoln}$

If $\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}=\\var{bc}$, then $y=$ [[0]].

Given $\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}=\\var{bc}$, we can rearrange the equation to get $y^\\frac{1}{\\var{bpower}}$ by itself and then we can raise both sides to the power of $\\var{bpower}$ to get $y$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}$	$=$	$\\var{bc}$

$\\simplify{{bxcoeff}y^(1/{bpower})}$	$=$	$\\simplify[basic]{{bc}-{bb}}$

$\\simplify{{bxcoeff}y^(1/{bpower})}$	$=$	$\\simplify{{bc-bb}}$

$y^\\frac{1}{\\var{bpower}}$	$=$	$\\simplify[!basic]{{bc-bb}/{bxcoeff}}$

$y^\\frac{1}{\\var{bpower}}$	$=$	$\\simplify{{bc-bb}/{bxcoeff}}$

$\\left(y^\\frac{1}{\\var{bpower}}\\right)^{\\var{bpower}}$	$=$	$\\simplify[basic]{({(bc-bb)/bxcoeff})^{bpower}}$

$y$	$=$	$\\var{bsoln}$

For this question, if the answer was $\\left(\\frac{35}{12}\\right)^{11}-24$, then you could enter (35/12)^(11)-24.

If $\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}=\\var{dc}$, then $z=$ [[0]].

Given $\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}=\\var{dc}$, we can rearrange the equation to get $\\simplify{(root(z+{db},{dpower}))}$ by itself, then we can raise both sides to the power of $\\var{dpower}$, and finally rearrange to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}$	$=$	$\\var{dc}$

$\\displaystyle{\\simplify{(root(z+{db},{dpower}))}}$	$=$	$\\simplify[basic]{{dc}*{ddenom}}$

$\\displaystyle{\\simplify{(root(z+{db},{dpower}))}}$	$=$	$\\var{dc*ddenom}$

$\\left(\\sqrt[\\var{dpower}]{\\simplify{z+{db}}}\\right)^\\var{dpower}$	$=$	$\\simplify[basic]{({dc*ddenom})^{dpower}}$

$\\simplify{z+{db}}$	$=$	$\\simplify[basic]{-{abs(dcddenom)}^{dpower}}$ $\\simplify[basic]{({abs(dcddenom)})^{dpower}}$ $\\simplify[basic]{({(dc*ddenom)})^{dpower}}$

$z$	$=$	$\\simplify[basic]{-{abs(dcddenom)}^{dpower}-{db}}$ $\\simplify[basic]{({abs(dcddenom)})^{dpower}-{db}}$ $\\simplify[basic]{({(dc*ddenom)})^{dpower}-{db}}$

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We can solve $\\var{a}x+\\var{b}=\\var{c}x$ by collecting like terms and rearranging for $x$.

This gives $x=$ [[0]].

Given $\\var{a}x+\\var{b}=\\var{c}x$, we can subtract $\\var{a}x$ from both sides to collect like terms, and then divide both sides by the coefficient of $x$ to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a}x+\\var{b}$	$=$	$\\var{c}x$

$\\var{a}x+\\var{b}-\\var{a}x$	$=$	$\\var{c}x-\\var{a}x$

$\\var{b}$	$=$	$\\var{c-a}x$

$\\displaystyle{\\frac{\\var{b}}{\\var{c-a}}}$	$=$	$\\displaystyle{\\frac{\\var{c-a}x}{\\var{c-a}}}$

$\\displaystyle{\\simplify{{b}/{c-a}}}$	$=$	$x$

$x$	$=$	$\\displaystyle{\\simplify{{b}/{c-a}}}$

There is often more than one way to solve an equation, one strategy used above in the first step was to get all the $x\\text{'s}$ one the side with the most $x\\text{'s}$, that way you end up with a positive number of $x\\text{'s}$. This is not necessary, we could have put all the $x$'s on the left-hand side but notice in this question that we then would have had to move the $\\var{b}$ onto the right-hand side, so it would have required more work, but nevertheless that method would result in the same result for $x$.

Solve $\\var{l}(\\var{m}w-\\var{n})=\\var{p}w+\\var{q}$ for $w$.

$w=$ [[0]]

Given $\\var{l}(\\var{m}w-\\var{n})=\\var{p}w+\\var{q}$, we can expand the brackets, get all the $w\\text{'s}$ on the left-hand side, collect like terms, get all the numbers on the right-hand side, collect like terms and then divide both sides by the coefficient of $w$ to get $w$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{l}(\\var{m}w-\\var{n})$	$=$	$\\var{p}w+\\var{q}$

$\\var{lm}w-\\var{nl}$	$=$	$\\var{p}w+\\var{q}$

$\\var{lm}w-\\var{nl}-\\var{p}w$	$=$	$\\var{p}w+\\var{q}-\\var{p}w$

$\\var{lm-p}w-\\var{nl}$	$=$	$\\var{q}$

$\\var{lm-p}w-\\var{nl}+\\var{n*l}$	$=$	$\\var{q}+\\var{n*l}$

$\\var{l*m-p}w$	$=$	$\\var{q+n*l}$

$\\displaystyle{\\frac{\\var{lm-p}w}{\\var{lm-p}}}$	$=$	$\\displaystyle{\\frac{\\var{q+nl}}{\\var{lm-p}}}$

$w$	$=$	$\\displaystyle{\\simplify{{q+nl}/{lm-p}}}$

Given $\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}=\\var{g}$, $y=$ [[0]].

Given $\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}=\\var{g}$, we can multiply both sides by $(y-\\var{f})$ to get rid of the fraction, get all the $y\\text{'s}$ on one side and the numbers on the other side, and then divide both sides by the coefficient of $y$ to get $y$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}$	$=$	$\\var{g}$

$\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}\\times(y-\\var{f})$	$=$	$\\var{g}\\times (y-\\var{f})$

$\\var{d}y$	$=$	$\\var{g}y+\\var{-g*f}$

$\\var{d}y+\\var{-g}y$	$=$	$\\var{g}y+\\var{-g*f}+\\var{-g}y$

$\\var{d-g}y$	$=$	$\\var{-g*f}$

$\\displaystyle{\\frac{\\var{d-g}y}{\\var{d-g}}}$	$=$	$\\displaystyle{\\frac{\\var{-g*f}}{\\var{d-g}}}$

$y$	$=$	$\\displaystyle{\\simplify{{-g*f}/{d-g}}}$

Solve $\\displaystyle{\\frac{z+\\var{h}}{z+\\var{j}}}=\\var{k}$ for $z$.

$z=$ [[0]]

Given $\\displaystyle{\\frac{z+\\var{h}}{z+\\var{j}}}=\\var{k}$, we can multiply both sides by $(z+\\var{j})$ to get rid of the fraction, get all the $z\\text{'s}$ on one side and the numbers on the other side, and then divide both sides by the coefficient of $z$ to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\frac{z+\\var{h}}{z+\\var{j}}}$	$=$	$\\var{k}$

$\\displaystyle{\\frac{z+\\var{h}}{z+\\var{j}}}\\times(z+\\var{j})$	$=$	$\\var{k}\\times (z+\\var{j})$

$z+\\var{h}$	$=$	$\\var{k}z+\\var{k*j}$

$z+\\var{h}-\\var{k}z$	$=$	$\\var{k}z+\\var{k*j}-\\var{k}z$

$\\var{1-k}z+\\var{h}$	$=$	$\\var{k*j}$

$\\var{1-k}z+\\var{h}-\\var{h}$	$=$	$\\var{k*j}-\\var{h}$

$\\var{1-k}z$	$=$	$\\var{k*j-h}$

$\\displaystyle{\\frac{\\var{1-k}z}{\\var{1-k}}}$	$=$	$\\displaystyle{\\frac{\\var{k*j-h}}{\\var{1-k}}}$

$z$	$=$	$\\displaystyle{\\simplify{({k*j-h})/({1-k})}}$

Solve $\\displaystyle{\\frac{x+\\var{add}}{\\var{denom1}}+\\frac{x}{\\var{denom2}}=\\var{right}}$.

$x=$ [[0]]

Given $\\displaystyle{\\frac{x+\\var{add}}{\\var{denom1}}+\\frac{x}{\\var{denom2}}=\\var{right}}$, we can multiply both sides by $\\var{denom1}$ and by $\\var{denom2}$ to get rid of the fractions, get all the $x\\text{'s}$ on one side and the numbers on the other side, and then divide both sides by the coefficient of $x$ to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\frac{x+\\var{add}}{\\var{denom1}}+\\frac{x}{\\var{denom2}}}$	$=$	$\\var{right}$

$\\displaystyle{\\left(\\frac{x+\\var{add}}{\\var{denom1}}\\right)\\times\\var{denom1}+\\left(\\frac{x}{\\var{denom2}}\\right)\\times\\var{denom1}}$	$=$	$\\var{right}\\times \\var{denom1}$	(multiply all terms by $\\var{denom1}$)

$\\displaystyle{x+\\var{add}+\\frac{\\var{denom1}x}{\\var{denom2}}}$	$=$	$\\var{r1}$

$\\displaystyle{(x+\\var{add})\\times\\var{denom2}+\\left(\\frac{\\var{denom1}x}{\\var{denom2}}\\right)\\times\\var{denom2}}$	$=$	$\\var{r1}\\times\\var{denom2}$	(multiply all terms by $\\var{denom2}$)

$\\displaystyle{\\var{denom2}x+\\var{a2}+\\var{denom1}x}$	$=$	$\\var{r12}$

$\\var{sumdeno}x+\\var{a2}$	$=$	$\\var{r12}$	(collect like terms)

$\\var{sumdeno}x$	$=$	$\\var{r12}-\\var{a2}$	(collect like terms)

$\\var{sumdeno}x$	$=$	$\\var{top}$

$x$	$=$	$\\displaystyle{\\simplify{{top}/({sumdeno})}}$	(divide by the coefficient of $x$)

Simplify the following by collecting like terms.

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$\\simplify[!collectnumbers]{{a[1]}x+{b[1]}y+{c[1]}z+{b[2]}y+{a[2]}x+{c[2]}z+{a[0]}x+{c[0]}z+{b[0]}y}$ = [[0]]

Like terms are terms where the variable part is the same. For example, $4x$ and $-x$ have the same variable part $x$. However, $3x$ and $-2y$ have different variable parts and are therefore unlike terms (or not like terms).

We can only collect like terms! Just like we can't say 2 m + 3 cm equals 5 m or 5 cm, we can't say $2x+3y$ equals $5x$ or $5y$! We can, however, say $2a+3a=5a$.

In our question we look at all the terms with a variable part of $x$ and add up all the corresponding coefficients, we do the same for the $y$ terms and the $z$ terms:

\\[\\begin{align}
&\\simplify[!collectnumbers]{{a[1]}x+{b[1]}y+{c[1]}z+{b[2]}y+{a[2]}x+{c[2]}z+{a[0]}x+{c[0]}z+{b[0]}y}\\\\
&=\\simplify[basic]{({a[1]}+{a[2]}+{a[0]})x+({b[1]}+{b[2]}+{b[0]})y+({c[1]}+{c[2]}+{c[0]})z}\\end{align}\\]

We present this as the sum of three unlike terms:

\\[\\simplify{{sum(a)}x+{sum(b)}y+{sum(c)}z}\\]

$\\simplify[!collectnumbers]{{d[1]}x^2+{f[1]}x+{g[1]}+{d[0]}x^2+{f[0]}x+{g[0]}}$ = [[0]]

Like terms are terms where the variable part is the same. For example, $4x$ and $-x$ have the same variable part $x$. However, $3x^2$ and $-2x$ have different variable parts and are therefore unlike terms (or not like terms).

We can only collect like terms! Just like we can't say 2 m + 3 cm equals 5 m or 5 cm, we can't say $2x^2+3x$ equals $5x^2$ or $5x$! We can, however, say $2x^2+3x^2=5x^2$.

In our question we look at all the terms with a variable part of $x^2$ and add up all the corresponding coefficients (the numbers in front of the variables), we do the same for the $x$ terms and the constant terms (the terms with no variable part):

\\[\\begin{align}&\\simplify[!collectnumbers]{{d[1]}x^2+{f[1]}x+{g[1]}+{d[0]}x^2+{f[0]}x+{g[0]}}\\\\&=\\simplify[basic]{({d[1]}+{d[0]})x^2+({f[1]}+{f[0]})x+({g[1]}+{g[0]})}\\end{align}\\]

We present this as the sum of three unlike terms:

\\[\\simplify[!noleadingminus, basic]{{sum(d)}x^2+{sum(f)}x+{sum(g)}}\\]

Expand and simplify the following.

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$\\simplify{(x+{a[0]})(x+{b[0]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{(x+{a[0]})(x+{b[0]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{x(x+{b[0]})+{a[0]}(x+{b[0]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{x^2+{b[0]}x+{a[0]}x+{a[0]*b[0]}}$

And collect like terms: $\\simplify[basic, unitfactor]{x^2+{a[0]+b[0]}x+{a[0]*b[0]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $x^2$, then the Outer terms to get $\\var{b[0]}x$, then the Inner terms to get $\\var{a[0]}x$, and then the Last terms to get $\\var{a[0]*b[0]}$. Now add them all together: $\\simplify[basic, unitfactor]{x^2+{a[0]+b[0]}x+{a[0]*b[0]}}$

Ensure you don't use brackets in your answer.

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$\\simplify{(x+{a[1]})(x+{b[1]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{(x+{a[1]})(x+{b[1]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{x(x+{b[1]})+{a[1]}(x+{b[1]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{x^2+{b[1]}x+{a[1]}x+{a[1]*b[1]}}$

And collect like terms: $\\simplify[basic, unitfactor]{x^2+{a[1]+b[1]}x+{a[1]*b[1]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $x^2$, then the Outer terms to get $\\var{b[1]}x$, then the Inner terms to get $\\var{a[1]}x$, and then the Last terms to get $\\var{a[1]*b[1]}$. Now add them all together: $\\simplify[basic, unitfactor]{x^2+{a[1]+b[1]}x+{a[1]*b[1]}}$

Ensure you don't use brackets in your answer.

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$\\simplify{(m+{a[2]})(m+{b[2]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{(m+{a[2]})(m+{b[2]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{m(m+{b[2]})+{a[2]}(m+{b[2]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{m^2+{b[2]}m+{a[2]}m+{a[2]*b[2]}}$

And collect like terms: $\\simplify[basic, unitfactor]{m^2+{a[2]+b[2]}m+{a[2]*b[2]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $m^2$, then the Outer terms to get $\\var{b[2]}m$, then the Inner terms to get $\\var{a[2]}m$, and then the Last terms to get $\\var{a[2]*b[2]}$. Now add them all together: $\\simplify[basic, unitfactor]{m^2+{a[2]+b[2]}m+{a[2]*b[2]}}$

Ensure you don't use brackets in your answer.

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$\\simplify{(t+{a[3]})(t+{b[3]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{(t+{a[3]})(t+{b[3]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{t(t+{b[3]})+{a[3]}(t+{b[3]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{t^2+{b[3]}t+{a[3]}t+{a[3]*b[3]}}$

And collect like terms: $\\simplify[basic, unitfactor]{t^2+{a[3]+b[3]}t+{a[3]*b[3]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $t^2$, then the Outer terms to get $\\var{b[3]}t$, then the Inner terms to get $\\var{a[3]}t$, and then the Last terms to get $\\var{a[3]*b[3]}$. Now add them all together: $\\simplify[basic, unitfactor]{t^2+{a[3]+b[3]}t+{a[3]*b[3]}}$

Ensure you don't use brackets in your answer.

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We start solving the equation one operation at a time by doing the inverse to both sides, when we get to undoing the log we recall the definition of $\\log_b$, write the equation in index form and continue solving.

Recall: The definition of $\\log_b$ says $\\log_b(a)=c$ is equivalent to $b^c=a$.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}})+\\var{f} }$	$=$	$\\var{g}$
$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}}) }$	$=$	$\\var{g-f}$	(subtract $\\var{f}$ from both sides)
$\\displaystyle{\\log_\\var{b}(\\simplify{{a}x+{c}}) }$	$=$	$\\var{power}$	(divide both sides by $\\var{d}$)
$\\simplify[basic]{{b}^{power}}$	$=$	$\\simplify{{a}x+{c}}$	(using the definition of $\\log_\\var{b}$)
$\\simplify[basic,unitpower]{{b}^{power}-{c}}$	$=$	$\\var{a}x$	(subtract $\\var{c}$ from both sides)

$\\displaystyle{\\simplify[basic,fractionnumbers,unitpower]{({b}^{power}-{c})/{a}}}$	$=$	$x$	(divide both sides by $\\var{a}$)

$x$	$=$	$\\displaystyle{\\simplify[basic,fractionnumbers,unitpower]{{({b}^{power}-{c})/{a}}}}$

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Solve the following equation for $x$

$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}})+\\var{f}=\\var{g} }.$

$x=$ [[0]]

Note: You can use the symbol ^ to signify powers, and / to signify division. Please ensure you use brackets correctly.

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m signifies 'minus'

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gcd(gtl,gtj)

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old version if(factorise(abs(ktf))[len(factorise(abs(ktf)))-1]=0,1,primes[len(factorise(abs(ktf)))-1])

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t signifies 'times'

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Find the $x$ and $y$ values that satisfy both of the following equations.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$y$	$=$	$\\simplify{{a}x+{b}}$	$(1)$
$y$	$=$	$\\simplify{{c}x+{d}}$	$(2)$

$x=$ [[0]], $y=$ [[1]]

There are many ways to solve these equations simultaneously. Here is one method.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$y$	$=$	$\\simplify{{a}x+{b}}$	$(1)$
$y$	$=$	$\\simplify{{c}x+{d}}$	$(2)$

Substitute the expression for $y$ given in $(1)$ into $(2)$:
\\[\\simplify{{a}x+{b} ={c}x+{d}}\\]

Collect like terms:
\\[\\simplify{{a-c}x={d-b}}\\]

Solve for $x$:
\\[x=\\var{xans}\\]

Now we know the $x$ value we can determine the corresponding $y$ value by substituting $x=\\var{xans}$ into either equation $(1)$ or $(2)$, below we substitute into $(1)$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$y$	$=$	$\\simplify[!collectnumbers]{{a}({xans})+{b}}$
	$=$	$\\var{yans}$

Therefore the values that satisfy equations $(1)$ and $(2)$ are $x=\\var{xans}$ and $y=\\var{yans}$.

Solve the following simultaneous equations.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{{f}x+{g}y}$	$=$	$\\var{-h}$	$(3)$
$\\simplify{{j}x+{k}y}$	$=$	$\\var{-l}$	$(4)$

$x=$ [[0]], $y=$ [[1]]

There are many ways to solve these equations simultaneously. Here is one method.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{{f}x+{g}y}$	$=$	$\\var{-h}$	$(3)$
$\\simplify{{j}x+{k}y}$	$=$	$\\var{-l}$	$(4)$

Solve one of the equations for one of the variables. Here we solve equation $(3)$ for $y$:

\\begin{align}\\var{g}y&=\\simplify{{-h}-{f}x}\\\\y&=\\simplify{({-h}-{f}x)/({g})}\\quad(5)\\end{align}

\\begin{align}y&=\\simplify{({-h}-{f}x)/({g})}\\quad(5)\\end{align}

Substitute this expression for $y$ given in $(5)$ into $(4)$:

\\[\\simplify[all,!collectnumbers]{{j}x+{k}(({-h}-{f}x)/({g})) = {-l}}\\]

Collect like terms:
\\[\\simplify[fractionnumbers]{{j-f*k/g}x={-l+h*k/g}}\\]

Solve for $x$:
\\[x=\\var{xbans}\\]

Now we know the $x$ value we can determine the corresponding $y$ value by substituting $x=\\var{xbans}$ into equation $(5)$:

\\begin{align}y&=\\simplify[unitdenominator,!collectnumbers]{({-h}-{f}({xbans}))/({g})}\\\\&=\\var{ybans}\\end{align}

Therefore the values that satisfy equations $(3)$ and $(4)$ are $x=\\var{xbans}$ and $y=\\var{ybans}$.

Solve the following system of equations.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{{m}a+{n}b+{p}}$	$=$	$0$	$(5)$
$\\simplify{{q}a+{r}b+{s}}$	$=$	$0$	$(6)$

$a=$ [[0]], $b=$ [[1]]

There are many ways to solve these equations simultaneously. Here is one method.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{{m}a+{n}b+{p}}$	$=$	$0$	$(5)$
$\\simplify{{q}a+{r}b+{s}}$	$=$	$0$	$(6)$

Solve one of the equations for one of the variables. Here we solve equation $(5)$ for $b$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{n}b$	$=$	$\\simplify{{-m}a+{-p}}$

$b$	$=$	$\\displaystyle{\\simplify{({-m}a+{-p})/({n})}}$	$(7)$

Substitute this expression for $b$ given in $(7)$ into $(6)$:

\\[\\simplify[all,!collectnumbers]{{q}a+{r}*(({-m}a+{-p})/{n}) + {s}=0}\\]

Collect like terms:
\\[\\simplify[fractionnumbers]{{q-r*m/n}a={-s+r*p/n}}\\]

Solve for $a$:
\\[a=\\simplify[fractionnumbers]{{xcans}}\\]

Now we know the $a$ value we can determine the corresponding $b$ value by substituting $a=\\simplify[fractionnumbers]{{xcans}}$ into equation $(7)$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$b$	$=$	$\\displaystyle{\\simplify[unitdenominator,!collectnumbers,fractionnumbers]{({-m}*({xcans})+{-p})/({n})}}$

	$=$	$\\simplify[fractionnumbers]{{ycans}}$

Therefore the values that satisfy equations $(5)$ and $(6)$ are $a=\\simplify[fractionnumbers]{{xcans}}$ and $b=\\simplify[fractionnumbers]{{ycans}}$.

Solve the following quadratic by factorisation:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$\\simplify{{aa}x^2-{bb}}$	$=$	$0$
$\\Longrightarrow$	[[0]]	$=$	0
$\\Longrightarrow$	$x$	$=$	[[1]]

Note: In the first gap, enter the quadratic in factored form.

Note: In the second gap, if $x=1,2$, enter set(1,2).

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "

Ensure you factorise the expression.

", "showStrings": false, "strings": ["xx", "x^2", "x**2"], "partialCredit": 0}, "vsetrangepoints": 5, "expectedvariablenames": ["x"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "answersimplification": "all", "scripts": {}, "answer": "({gg})({a/g}x+{b/g})({a/g}x-{b/g})", "marks": 1, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme", "musthave": {"message": "

Ensure you factorise the expression.

", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}}, {"vsetrangepoints": 5, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "answersimplification": "all", "scripts": {}, "answer": "set({b}/{a},-{b}/{a})", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "steps": [{"prompt": "

Since $\\simplify{{aa}x^2}$ is $\\simplify{{a}x}$ squared and $\\var{bb}$ is $\\var{b}$ squared, we can recognise $\\simplify{{aa}x^2-{bb}}$ as a difference of two squares.

Recalling that $(a+b)(a-b)=a^2-b^2$, we have

$\\simplify{{aa}x^2-{bb}}=(\\simplify{{a}x+{b}})(\\simplify{{a}x-{b}})$

Now, by the null factor law, either

$\\simplify{{a}x+{b}}=0$ or $\\simplify{{a}x-{b}}=0$.

Solving these equations results in

$x=\\simplify{-{b}/{a}}$ or $x=\\simplify{{b}/{a}}$.

Notice there is a common factor of $\\var{gg}$ that we can deal with first

$\\simplify{{aa}x^2-{bb}}=\\simplify{{gg}({aa/gg}x^2-{bb/gg})}.$

Next, notice the remaining expression is a difference of two squares. Recalling that $(a+b)(a-b)=a^2-b^2$, we have

$\\simplify{{aa}x^2-{bb}}=\\simplify{{gg}({aa/gg}x^2-{bb/gg})}=\\simplify{({gg})({a/g}x+{b/g})({a/g}x-{b/g})}$

Now, by the null factor law, either

$\\simplify{{a/g}x+{b/g}}=0$ or $\\simplify{{a/g}x-{b/g}}=0$.

Solving these equations results in

$x=\\simplify{-{b}/{a}}$ or $x=\\simplify{{b}/{a}}$.

Solve the following quadratic by factorisation:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$\\simplify{{c[0]}^2/{c[2]}^2 x^2-{c[1]}^2/{c[3]}^2}$	$=$	0
$\\Longrightarrow$	[[0]]	$=$	0
$\\Longrightarrow$	$x$	$=$	[[1]]

Note: In the first gap, enter the quadratic in factored form.

Note: In the second gap, if $x=1,2$, enter set(1,2).

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "

Ensure you factorise the expression.

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Ensure you factorise the expression.

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We should recognise this as a difference of two squares, where $\\simplify{{c[0]}^2/{c[2]}^2 x^2}$ is $\\left(\\simplify{{c[0]}/{c[2]}x}\\right)^2$ and $\\simplify{{c[1]}^2/{c[3]}^2}$ is $\\left(\\simplify{{c[1]}/{c[3]}}\\right)^2$. Therefore

$\\simplify{{c[0]}^2/{c[2]}^2 x^2-{c[1]}^2/{c[3]}^2}=\\simplify{({c[0]}/{c[2]}x+{c[1]}/{c[3]})({c[0]}/{c[2]}x-{c[1]}/{c[3]})}.$

Now, by the null factor law, either

$\\simplify{{c[0]}/{c[2]}x+{c[1]}/{c[3]}}=0$ or $\\simplify{{c[0]}/{c[2]}x-{c[1]}/{c[3]}}=0$.

Solving these equations results in

$x=\\simplify{-{c[1]*c[2]}/{c[3]*c[0]}}$ or $x=\\simplify{{c[1]*c[2]}/{c[3]*c[0]}}$.

We aim to solve the quadratic $\\simplify{x^2+{linear}x+{const}}$ by factorisation.

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The quadratic $\\simplify{x^2+{linear}x+{const}}=0$ factorised is [[0]]$\\;=0$

Note: In the gap above, enter the quadratic in factored form.

Since $(x+a)(x+b)=x^2+(a+b)x+ab$, when we are factorising a quadratic, such as $x^2+cx+d$, we must find the numbers $a$ and $b$ such that $c=a+b$ and $d=ab$.

In the case of $\\simplify{x^2+{linear}x+{const}}$ we ask

what two numbers add to give $\\var{linear}$ and multiply to give $\\var{const}$?

Therefore the numbers must be $\\var{a}$ and $\\var{b}$, that is

$\\simplify{x^2+{linear}x+{const}}=(\\simplify{x+{a}})(\\simplify{x+{b}}).$

You can check this by expanding the binomial product.

Ensure you factorise the expression.

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Ensure you factorise the expression.

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The above factorisation and the null factor law implies $x=$ [[0]], [[1]].

Now, using the null factor law we have, either $\\simplify{x+{a}}=0$ or $\\simplify{x+{b}}=0$. In otherwords, either $x=\\var{-a}$ or $\\var{-b}$.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$\\simplify{x^2+{linear}x+{const}}$	$=$	0
$\\Longrightarrow$	$ (\\simplify{x+{a}})(\\simplify{x+{b}})$	$=$	0
$\\Longrightarrow$	$x$	$=$	$\\var{-a},\\var{-b}$

Solve the following quadratic by factorisation:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$\\simplify{{aa}x^2+{mid}x+{bb}}$	$=$	0
$\\Longrightarrow$	[[0]]	$=$	0
$\\Longrightarrow$	$x$	$=$	[[1]]

Note: In the first gap, enter the quadratic in factored form.

Note: In the second gap, there should only be one solution.

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Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.

In fact, $\\simplify{{aa}x^2+{mid}x+{bb}}$ is also a perfect square, since $\\var{aa}x^2=(\\var{a}x)^2$, $\\var{bb}=\\simplify{({b})^2}$, and $\\var{mid}x=2(\\var{a}x)(\\var{b})$.

That is, $\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{({a}x+{b})^2}$.

Now, since $\\simplify{({a}x+{b})^2}=0$ we can take the (plus or minus) square root of both sides to get $\\simplify{({a}x+{b})}=0$. We can then solve this for $x$ to find $x=\\simplify{-{b}/{a}}$.

Each bracket has a common factor of $\\var{g}$, so we can move both of them to the front, to write

$\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{{gg}({a/g}x+{b/g})^2}$.

Now, using the null factor law, since $\\simplify{{gg}({a/g}x+{b/g})^2}=0$ we must have $\\simplify{{a/g}x+{b/g}}=0$. We can then solve this for $x$ to find $x=\\simplify{{-b}/{a}}$.

Ensure you factorise the expression.

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Ensure you factorise the expression.

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Solve the following quadratic by factorisation:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]d}/{(c[1]c[2])}x+{d^2}/{c[2]^2}}$	$=$	0
$\\Longrightarrow$	[[0]]	$=$	0
$\\Longrightarrow$	$x$	$=$	[[1]]

Note: In the first gap, enter the quadratic in factored form.

Note: In the second gap, there should only be one solution.

Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.

In fact, $\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ is also a perfect square, since

$\\simplify{{c[0]^2}/{c[1]^2}}x^2=\\left(\\simplify{{c[0]}/{c[1]}}x\\right)^2$,
$\\simplify{{d^2}/{c[2]^2}}=\\left(\\simplify{{d}/{c[2]}}\\right)^2$ and
$\\simplify{{2c[0]*d}/{(c[1]*c[2])}x}=2\\left(\\simplify{{c[0]}/{c[1]}x}\\right)\\left(\\simplify{{d}/{c[2]}}\\right)$.

That is,

$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}=\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}$.

Now, since $\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}=0$ we can take the (plus or minus) square root of both sides to get $\\simplify{{c[0]}/{c[1]}x+{d}/{c[2]}}=0$. We can then solve this for $x$ to find $x=\\simplify{{-d*c[1]}/{c[0]*c[2]}}$.

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Ensure you factorise the expression.

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Ensure you factorise the expression.

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Given $x+\\var{a}=\\var{b}$, we can rearrange the equation to that find $x=$ [[0]].

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Given $x+\\var{a}=\\var{b}$, we subtract $\\var{a}$ from both sides to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$x+\\var{a}$	$=$	$\\var{b}$
$x+\\var{a}-\\var{a}$	$=$	$\\var{b}-\\var{a}$
$x$	$=$	$\\var{ans1}$

Given $y-\\var{c}=\\var{d}$, $y=$ [[0]].

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Given $y-\\var{c}=\\var{d}$, we add $\\var{c}$ to both sides to get $y$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$y-\\var{c}$	$=$	$\\var{d}$
$y-\\var{c}+\\var{c}$	$=$	$\\var{d}+\\var{c}$
$y$	$=$	$\\var{ans2}$

Rearrange $\\var{f}+z=\\var{g}$ to determine the value of $z$.

$z=$ [[0]]

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Given $\\var{f}+z=\\var{g}$, we subtract $\\var{f}$ from both sides to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{f}+z$	$=$	$\\var{g}$
$\\simplify[basic]{{f}+z-{f}}$	$=$	$\\simplify[basic]{{g}-{f}}$
$z$	$=$	$\\var{ans3}$

Solve $\\var{h}=\\var{j}+a$ for $a$.

$a=$ [[0]]

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Given $\\var{h}=\\var{j}+a$, we subtract $\\var{j}$ from both sides to get $a$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{h}$	$=$	$\\var{j}+a$
$\\simplify[basic]{{h}-{j}}$	$=$	$\\simplify[basic]{{j}+a-{j}}$
$\\var{ans4}$	$=$	$a$
$a$	$=$	$\\var{ans4}$

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Given $\\var{a}x=\\var{b}$, we can rearrange the equation to that find $x=$ [[0]].

Given $\\var{a}x=\\var{b}$, we divide both sides by $\\var{a}$ to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a}x$	$=$	$\\var{b}$

$\\displaystyle{\\frac{\\var{a}x}{\\var{a}}}$	$=$	$\\displaystyle{\\frac{\\var{b}}{\\var{a}}}$

$x$	$=$	$\\displaystyle{\\simplify{{b}/{a}}}$

Given $\\var{c}y=\\var{d}$, $y=$ [[0]].

Given $\\var{c}y=\\var{d}$, we divide both sides by $\\var{c}$ to get $y$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{c}y$	$=$	$\\var{d}$

$\\displaystyle{\\frac{\\var{c}y}{\\var{c}}}$	$=$	$\\displaystyle{\\frac{\\var{d}}{\\var{c}}}$

$y$	$=$	$\\displaystyle{\\simplify{{d}/{c}}}$

Rearrange $\\displaystyle{\\frac{z}{\\var{f}}}=\\var{g}$ to determine the value of $z$.

$z=$ [[0]]

Given $\\displaystyle{\\frac{z}{\\var{f}}}=\\var{g}$, we multiply both sides by $\\var{f}$ to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\frac{z}{\\var{f}}}$	$=$	$\\var{g}$

$\\displaystyle{\\frac{z}{\\var{f}}}\\times \\var{f}$	$=$	$\\var{g}\\times\\var{f}$

$z$	$=$	$\\var{ans3}$

Solve $\\displaystyle{\\var{h}=-\\frac{a}{\\var{j}}}$ for $a$.

$a=$ [[0]]

Given $\\displaystyle{\\var{h}=-\\frac{a}{\\var{j}}}$, we multiply both sides by $-\\var{j}$ to get $a$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{h}$	$=$	$\\displaystyle{-\\frac{a}{\\var{j}}}$

$\\var{h}\\times(-\\var{j})$	$=$	$\\displaystyle{-\\frac{a}{\\var{j}}\\times(-\\var{j})}$

$\\var{ans4}$	$=$	$a$

$a$	$=$	$\\var{ans4}$

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Given $\\var{a}x+\\var{b}=\\var{c}$, solving for $x$ gives $x=$ [[0]].

Given $\\var{a}x+\\var{b}=\\var{c}$, we can subtract $\\var{b}$ from both sides to get $\\var{a}x$ by itself, and then divide both sides by $\\var{a}$ to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a}x+\\var{b}$	$=$	$\\var{c}$

$\\var{a}x+\\var{b}-\\var{b}$	$=$	$\\var{c}-\\var{b}$

$\\var{a}x$	$=$	$\\var{c-b}$

$\\displaystyle{\\frac{\\var{a}x}{\\var{a}}}$	$=$	$\\displaystyle{\\frac{\\var{c-b}}{\\var{a}}}$

$x$	$=$	$\\displaystyle{\\simplify{{c-b}/{a}}}$

Given $\\var{d}-\\var{f}y=\\var{g}$, $y=$ [[0]].

Given $\\var{d}-\\var{f}y=\\var{g}$, we can subtract $\\var{d}$ from both sides to get $-\\var{f}y$ by itself, and then divide both sides by $-\\var{f}$ to get $y$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{d}-\\var{f}y$	$=$	$\\var{g}$

$\\var{d}-\\var{f}y-\\var{d}$	$=$	$\\var{g}-\\var{d}$

$-\\var{f}y$	$=$	$\\var{g-d}$

$\\displaystyle{\\frac{\\var{-f}y}{\\var{-f}}}$	$=$	$\\displaystyle{\\frac{\\var{g-d}}{\\var{-f}}}$

$y$	$=$	$\\displaystyle{\\simplify{{g-d}/{-f}}}$

Rearrange $\\displaystyle{\\frac{z}{\\var{h}}}-\\var{j}=\\var{k}$ to determine the value of $z$.

$z=$ [[0]]

Given $\\displaystyle{\\frac{z}{\\var{h}}}-\\var{j}=\\var{k}$, we add $\\var{j}$ to both sides to get $\\displaystyle{\\frac{z}{\\var{h}}}$ by itself and then multiply both sides by $\\var{h}$ to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\frac{z}{\\var{h}}}-\\var{j}$	$=$	$\\var{k}$

$\\displaystyle{\\frac{z}{\\var{h}}}-\\var{j}+\\var{j}$	$=$	$\\var{k}+\\var{j}$

$\\displaystyle{\\frac{z}{\\var{h}}}$	$=$	$\\var{k+j}$

$\\displaystyle{\\frac{z}{\\var{h}}\\times\\var{h}}$	$=$	$\\var{k+j}\\times \\var{h}$

$z$	$=$	$\\var{ans3}$

Solve $\\displaystyle{\\frac{a-\\var{l}}{\\var{m}}}=\\var{n}$ for $a$.

$a=$ [[0]]

Given $\\displaystyle{\\frac{a-\\var{l}}{\\var{m}}}=\\var{n}$, we can multiply both sides by $\\var{m}$ to get $a-\\var{l}$ by itself and then add $\\var{l}$ to both sides to get $a$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\frac{a-\\var{l}}{\\var{m}}}$	$=$	$\\var{n}$

$\\displaystyle{\\frac{a-\\var{l}}{\\var{m}}}\\times \\var{m}$	$=$	$\\var{n}\\times\\var{m}$

$a-\\var{l}$	$=$	$\\var{n*m}$

$a-\\var{l}+\\var{l}$	$=$	$\\var{n*m}+\\var{l}$

$a$	$=$	$\\var{ans4}$

Solve $\\var{p}=\\var{q}(\\var{r}+b)$.

$b=$ [[0]]

Given $\\var{p}=\\var{q}(\\var{r}+b)$, we can divide both sides by $\\var{q}$ to get $\\var{r}+b$ by itself and then subtract $\\var{r}$ from both sides to get $b$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{p}$	$=$	$\\var{q}(\\var{r}+b)$

$\\displaystyle{\\frac{\\var{p}}{\\var{q}}}$	$=$	$\\displaystyle{\\frac{\\var{q}(\\var{r}+b)}{\\var{q}}}$

$\\displaystyle{\\simplify{{p}/{q}}}$	$=$	$\\var{r}+b$

$\\displaystyle{\\simplify{{p}/{q}}}-\\var{r}$	$=$	$\\var{r}+b-\\var{r}$

$\\displaystyle{\\simplify{{p-r*q}/{q}}}$	$=$	$b$

$b$	$=$	$\\displaystyle{\\simplify{{p-r*q}/{q}}}$

Solve $\\displaystyle{\\frac{\\var{s}w}{\\var{t}}}=\\var{u}$.

$w=$ [[0]]

Given $\\displaystyle{\\frac{\\var{s}w}{\\var{t}}}=\\var{u}$, we can multiply both sides by $\\var{t}$ to get $\\var{s}w$ by itself and then divide both sides by $\\var{s}$ to get $w$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle{\\frac{\\var{s}w}{\\var{t}}}$	$=$	$\\var{u}$

$\\displaystyle{\\frac{\\var{s}w}{\\var{t}}}\\times\\var{t}$	$=$	$\\var{u}\\times\\var{t}$

$\\var{s}w$	$=$	$\\var{u*t}$

$\\displaystyle{\\frac{\\var{s}w}{\\var{s}}}$	$=$	$\\displaystyle{\\frac{\\var{u*t}}{\\var{s}}}$

$w$	$=$	$\\displaystyle{\\simplify{{u*t}/{s}}}$

Given that $\\displaystyle{(\\simplify{x+{a}})(\\simplify{{b}x+{c}})=0}$. Determine the set of possible values of $x$.

$x=$ [[0]]

Note: if your answer is $1$ and $2$ input set(1,2)

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Notice, this is a quadratic that has already been factorised. You don't need to expand it since we have the null factor law:

\\[\\text{If } ab=0, \\text{ then } a=0 \\text{ or } b=0.\\]

Since

\\[(\\simplify{x+{a}})(\\simplify{{b}x+{c}})=0,\\]

this means $\\simplify{x+{a}}=0$ or $\\simplify{{b}x+{c}}=0$. Solving each of these equations gives $x=\\var{-a}$ or $x=\\simplify{-{c}/{b}}$.

For this question, we input our answer as set$\\left(\\var{-a},\\simplify{-{c}/{b}}\\right)$.

Solve $\\displaystyle{\\simplify{{l}a}(\\simplify{a+{d}})(\\simplify{{f}a+{g}})\\left(\\simplify{{h}a+{j}/{k}}\\right)\\left(\\simplify{{m}a/{n}+{p}/{q}}\\right)=0}$ for $a$.

$a=$ [[0]]

Note: if your answer is $1$, $2$ and $3$ input set(1,2,3)

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Notice, this expression that has already been factorised. You don't need to expand it since we have the null factor law:

\\[\\text{If } ab=0, \\text{ then } a=0 \\text{ or } b=0.\\]

Since

\\[\\simplify{{l}a}(\\simplify{a+{d}})(\\simplify{{f}a+{g}})\\left(\\simplify{{h}a+{j}/{k}}\\right)\\left(\\simplify{{m}a/{n}+{p}/{q}}\\right)=0,\\]

this means $\\simplify{{l}a}=0$, $\\simplify{a+{d}}=0$, $\\simplify{{f}a+{g}}=0$, $\\simplify{{h}a+{j}/{k}}=0$, or $\\simplify{{m}a/{n}+{p}/{q}}=0$. Solving each of these equations gives $x=0$, $x=\\var{-d}$, $x=\\var[fractionnumbers]{-g/f}$, $x=\\var[fractionnumbers]{-j/(k*h)}$, or $x=\\var[fractionnumbers]{(-p*n)/(q*m)}$.

For this question, we input our answer as set$\\left(0,\\var{-d},\\var[fractionnumbers]{-g/f},\\var[fractionnumbers]{-j/(k*h)},\\var[fractionnumbers]{(-p*n)/(q*m)}\\right)$.

Evaluate the following, without the use of a calculator:

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For an insight into an index of 0, consider the following table:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

index form	$b^3$	$b^2$	$b^1$	$b^0$
expanded form	$b\\times b \\times b$	$b \\times b$	$b$	[[0]]

Notice each time the power decreases by $1$, the result is divided by $b$. Using this idea, fill in the rest of the table.

Each time you reduce the power by 1 you divide the result by the base, that is $b$. Following this pattern:

\\[b^{0}=\\frac{b}{b}=1\\]

From this we conclude that any non-zero number to the power of 0 is 1.

$x^0$ = [[0]]

Any non-zero number to the power of 0 is 1.

$(\\var{coeff1}r)^0$ = [[0]]

Any non-zero number to the power of 0 is 1.

$-t^0$ = [[0]]

Here $t$ is to the power of zero and so it evaluates to $1$. However, there is still a negative sign in front, so the answer must be $-1$.

Note $(-t)^0$ is quite different to $-t^0$.

$\\displaystyle\\left(\\frac{\\var{coeff2}y}{\\var{coeff3}x}\\right)^0$ = [[0]]

Any non-zero number to the power of 0 is 1.

$\\displaystyle\\frac{m^0}{z}$ = [[0]]

Note that the power of zero is only acting on the numerator. As such the numerator is equal to $1$ and the denominator remains untouched.

$(r \\times s)^0$ = [[0]]

Regardless of the result of the multiplication the power of zero will force it to evaluate to 1.

$d\\times e^0$ = [[0]]

The power of zero is only acting on the second number in the product. It is only this number that will become 1.

Simplify the following

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$\\displaystyle \\frac{\\sqrt[\\var{root}]{ a^\\var{n}\\times b\\times a^\\var{root-n}}}{ c^\\var{-m}\\times a\\times b^0\\times b^{1/\\var{root}}}\\times \\left(\\frac{ a^\\var{p}\\times b}{ c}\\right)^\\var{m}+\\var{mult}( a^\\var{root-n}\\times b^\\var{m+root}\\times c^\\var{n+m})^0-\\frac{ a^\\var{p*m-n}\\times a^\\var{n}}{ b^\\var{-m}}$ = [[0]]

The order shown is simply one order in which to do things, you also don't need to use as many lines as are shown here. There are only three terms in this expression. We will simplify them individually and then combine them.

The middle term $\\var{mult}(a^\\var{root-n}\\times b^\\var{m+root}\\times c^\\var{n+m})^0$ has a power of zero acting on a bracket so this can be reduced to simply $\\var{mult}\\times 1=\\var{mult}$.

The third term is slightly more complicated in that we will need to add indices (because we are multiplying expressions with the same base) and deal with negative indices:

\\begin{align}&\\frac{a^\\var{p*m-n}\\times a^\\var{n}}{b^\\var{-m}}\\\\
&=\\frac{a^\\color{red}{\\var{p*m}}}{b^\\var{-m}}\\\\
&=a^\\var{p*m}\\times\\color{red}{b^\\var{m}}\\end{align}

The first term is more complicated and we will use all of our index laws to simplify it:
\\begin{align}&\\frac{\\sqrt[\\var{root}]{a^\\var{n}\\times b\\times a^\\var{root-n}}}{c^\\var{-m}\\times a\\times b^0\\times b^{1/\\var{root}}}\\times \\left(\\frac{a^\\var{p}\\times b}{c}\\right)^\\var{m}\\\\
&=\\frac{\\left(a^\\var{n}\\times b\\times a^\\var{root-n}\\right)^{\\color{red}{1/\\var{root}}}}{c^\\var{-m}\\times a\\times \\color{red}{1}\\times b^{1/\\var{root}}}\\times \\left(\\frac{a^\\color{red}{\\var{p*m}}\\times b^\\color{red}{\\var{m}}}{c^\\color{red}{\\var{m}}}\\right)\\\\
&=\\frac{\\left(a^\\color{red}{\\var{root}}\\times b\\right)^{1/\\var{root}}\\color{red}{\\times}a^\\var{p*m}\\times b^\\var{m}}{c^\\var{-m}\\times a\\times b^{1/\\var{root}}\\color{red}{\\times} c^\\var{m}}\\\\
&=\\frac{a^\\color{red}{1}\\times b^\\color{red}{1/\\var{root}}\\color{red}{\\times}a^\\var{p*m}\\times b^\\var{m}}{c^\\color{red}{\\var{0}}\\times a\\times b^{1/\\var{root}}}\\\\
&=\\frac{a^\\color{red}{\\var{1+p*m}}\\times b^\\color{red}{\\var{m}+1/\\var{root}}}{\\color{red}{1}\\times a\\times b^{1/\\var{root}}}\\\\
&=a^\\color{red}{\\var{p*m}}\\times b^\\color{red}{\\var{m}}\\\\
\\end{align}

This is the same as the third term.

So far we have reduced our original expression to $a^\\color{red}{\\var{p*m}}\\times b^\\color{red}{\\var{m}}+\\var{mult}-a^\\color{red}{\\var{p*m}}\\times b^\\color{red}{\\var{m}}$ but this is simply $\\var{mult}$.

Your answer is longer than necessary.

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Expand and simplify the following.

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$\\simplify{(x+{a[0]})(x-{a[0]})}$ = [[0]]

Method 1 (difference of two squares)

Notice that the product will expand to be a difference of two squares. Therefore, we can square the first term and subtract the square of the second term (see the other methods below if you aren't sure why):

$\\simplify{(x+{a[0]})(x-{a[0]})}=\\simplify{x^2-{a[0]*a[0]}}$

Method 2 (the distributive law)

We expand $\\simplify[basic]{(x+{a[0]})(x-{a[0]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{x(x-{a[0]})+{a[0]}(x-{a[0]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{x^2-{a[0]}x+{a[0]}x-{a[0]*a[0]}}$

And collect like terms: $\\simplify[basic, unitfactor]{x^2-{a[0]*a[0]}}$

Method 3 (FOIL)

Multiply the First terms in each bracket to get $x^2$, then the Outer terms to get $\\var{-a[0]}x$, then the Inner terms to get $\\var{a[0]}x$, and then the Last terms to get $-\\var{a[0]*a[0]}$. Now add them all together: $\\simplify[basic, unitfactor]{x^2-{a[0]*a[0]}}$

Ensure you don't use brackets in your answer.

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$\\simplify{(r+{a[3]})(r+{a[3]})}$ = [[0]]

Method 1 (perfect square)

Notice that $\\simplify{(r+{a[3]})(r+{a[3]})}$ is a perfect square. Therefore, we can square the first term $r$, double the product of the two terms $r$ and $\\var{a[3]}$, then square the last term $\\var{a[3]}$, add them all together (see the other methods below if you aren't sure why):

$\\simplify{(r+{a[3]})(r+{a[3]})}=\\simplify{r^2+{2*a[3]}r+{a[3]*a[3]}}$

Method 2 (the distributive law)

We expand $\\simplify[basic]{(r+{a[3]})(r+{a[3]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{r(r+{a[3]})+{a[3]}(r+{a[3]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{r^2+{a[3]}r+{a[3]}r+{a[3]*a[3]}}$

And collect like terms: $\\simplify[basic, unitfactor]{r^2+{a[3]+a[3]}r+{a[3]*a[3]}}$

Method 3 (FOIL)

Multiply the First terms in each bracket to get $r^2$, then the Outer terms to get $\\var{a[3]}r$, then the Inner terms to get $\\var{a[3]}r$, and then the Last terms to get $\\var{a[3]*a[3]}$. Now add them all together: $\\simplify[basic, unitfactor]{r^2+{a[3]+a[3]}r+{a[3]*a[3]}}$

Ensure you don't use brackets in your answer.

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$\\simplify{(x+{a[2]})^2}$ = [[0]]

It is important to realise that $\\simplify{(x+{a[2]})^2}=\\simplify[basic]{(x+{a[2]})(x+{a[2]})}$. Recall that squaring something is multiplying it by itself.

Method 1 (perfect square)

Notice that $\\simplify{(x+{a[3]})(x+{a[3]})}$ is a perfect square. Therefore, we can square the first term $r$, double the product of the two terms $r$ and $\\var{a[3]}$, then square the last term $\\var{a[3]}$, add them all together (see the other methods below if you aren't sure why):

$\\simplify{(x+{a[3]})(x+{a[3]})}=\\simplify{x^2+{2*a[3]}x+{a[3]*a[3]}}$

Method 2 (the distributive law)

We expand $\\simplify[basic]{(x+{a[3]})(x+{a[3]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{x(x+{a[3]})+{a[3]}(x+{a[3]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{x^2+{a[3]}x+{a[3]}x+{a[3]*a[3]}}$

And collect like terms: $\\simplify[basic, unitfactor]{x^2+{a[3]+a[3]}x+{a[3]*a[3]}}$

Method 3 (FOIL)

Multiply the First terms in each bracket to get $x^2$, then the Outer terms to get $\\var{a[3]}x$, then the Inner terms to get $\\var{a[3]}x$, and then the Last terms to get $\\var{a[3]*a[3]}$. Now add them all together: $\\simplify[basic, unitfactor]{x^2+{a[3]+a[3]}x+{a[3]*a[3]}}$

Ensure you don't use brackets in your answer.

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$\\simplify{{prime^primepower}x^{xpower}y^{ypower}(x+{xconstant})^{xconstantpower} (z+{failconstant})^{failsafepower}}$

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Which of the following are factors of $\\simplify{{prime^primepower}x^{xpower}y^{ypower}(x+{xconstant})^{xconstantpower} (z+{failconstant})^{failsafepower}}$?

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Factors are things that multiply to make a product.

Consider the product $\\simplify{{prime^primepower}x^{xpower}y^{ypower}(x+{xconstant})^{xconstantpower} (z+{failconstant})^{failsafepower}}$.

We can write this product as

$\\simplify[basic, alwaystimes]{{expanded}}$

Any combination of the above factors will still be a factor (since we can rearrange the product so that our collection of factors are all together and we can treat that collection as a single factor).

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Suppose that $\\var{factor1}${factor2} is a factor of an expression. What can be said of $-\\var{factor1}${factor2}?

Since $-1$ can divide every number (just like $1$ can), if a positive number is a factor, then so is the negative of it.

In particular, if $\\var{factor1}${factor2} is a factor of an expression, then that expression can be written as a product involving $\\var{factor1}${factor2}, for instance $\\var{factor1}${factor2}$w$ where $w$ is the other factor. Notice then that we can write the expression as $-\\var{factor1}${factor2}$(-w)$, that is, $-\\var{factor1}${factor2} is also a factor.

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It is also a factor.

", "

It is not necessarily a factor.

", "

It is definitely not a factor.

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$\\simplify{{aa}x^2-{bb}}$ = [[0]].

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Ensure you factorise the expression.

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Since $\\simplify{{aa}x^2}$ is $\\simplify{{a}x}$ squared and $\\var{bb}$ is $\\var{b}$ squared, we can recognise $\\simplify{{aa}x^2-{bb}}$ as a difference of two squares.

Recalling that $(a+b)(a-b)=a^2-b^2$, we have

$\\simplify{{aa}x^2-{bb}}=(\\simplify{{a}x+{b}})(\\simplify{{a}x-{b}})$

Notice there is a common factor of $\\var{gg}$ that we can deal with first

$\\simplify{{aa}x^2-{bb}}=\\simplify{{gg}({aa/gg}x^2-{bb/gg})}.$

Next, notice the remaining expression is a difference of two squares. Recalling that $(a+b)(a-b)=a^2-b^2$, we have

$\\simplify{{aa}x^2-{bb}}=\\simplify{{gg}({aa/gg}x^2-{bb/gg})}=\\simplify{({gg})({a/g}x+{b/g})({a/g}x-{b/g})}$

$\\simplify{{c[0]}^2/{c[2]}^2 x^2-{c[1]}^2/{c[3]}^2}$ = [[0]].

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Ensure you factorise the expression.

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$\\simplify{{c[0]}^2/{c[2]}^2 x^2-{c[1]}^2/{c[3]}^2}=\\simplify{({c[0]}/{c[2]}x+{c[1]}/{c[3]})({c[0]}/{c[2]}x-{c[1]}/{c[3]})}.$

Factorise the following into linear factors. That is, write the quadratic as a product of terms that look like $ax+b$ where $a$ and $b$ are real numbers.

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Factorise the following into linear factors. That is, write the quadratic as a product of terms that look like $ax+b$ where $a$ and $b$ are real numbers.

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$\\simplify{x^2+{linear}x+{const}}$ = [[0]].

Since $(x+a)(x+b)=x^2+(a+b)x+ab$, when we are factorising a quadratic, such as $x^2+cx+d$, we must find the numbers $a$ and $b$ such that $c=a+b$ and $d=ab$.

In the case of $\\simplify{x^2+{linear}x+{const}}$ we ask

what two numbers add to give $\\var{linear}$ and multiply to give $\\var{const}$?

Therefore the numbers must be $\\var{a}$ and $\\var{b}$, that is

$\\simplify{x^2+{linear}x+{const}}=(\\simplify{x+{a}})(\\simplify{x+{b}}).$

You can check this by expanding the binomial product.

Ensure you factorise the expression.

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Ensure you factorise the expression.

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Factorise the following into squares, that is, $(ax+b)^2$ or if there is a common factor $k$, then $k(cx+d)^2$.

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$\\simplify{{aa}x^2+{mid}x+{bb}}$ = [[0]].

Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.

In fact, $\\simplify{{aa}x^2+{mid}x+{bb}}$ is also a perfect square, since

$\\var{aa}x^2=(\\var{a}x)^2$,
$\\var{bb}=\\simplify{({b})^2}$ and
$\\var{mid}x=2(\\var{a}x)(\\var{b})$.

That is, $\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{({a}x+{b})^2}$.

Each bracket has a common factor of $\\var{g}$, so we call move both of them to the front, to get

$\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{{gg}({a/g}x+{b/g})^2}$.

You can always check your factorisation by expanding.

Ensure you factorise the expression.

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Ensure you factorise the expression.

$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ = [[0]].

Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.

In fact, $\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ is also a perfect square, since

$\\simplify{{c[0]^2}/{c[1]^2}}x^2=\\left(\\simplify{{c[0]}/{c[1]}}x\\right)^2$,
$\\simplify{{d^2}/{c[2]^2}}=\\left(\\simplify{{d}/{c[2]}}\\right)^2$ and
$\\simplify{{2c[0]*d}/{(c[1]*c[2])}x}=2\\left(\\simplify{{c[0]}/{c[1]}x}\\right)\\left(\\simplify{{d}/{c[2]}}\\right)$.

That is,

$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}=\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}$.

You can always check your factorisation by expanding.

Ensure you factorise the expression.

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Ensure you factorise the expression.

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The expression $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ is a sum and can be factorised (written as a product) by finding the largest common factor:

$\\var{pmult*pxcoeff}x+\\var{pconstant*pmult} = $ [[0]] $\\large($ [[1]] $\\large)$

We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

The (largest) common factor of $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ is $\\var{pmult}$.

Once we remove that factor from each term in $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ we are left with $\\var{pxcoeff}x+\\var{pconstant}$.

That means $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}= \\var{pmult}(\\var{pxcoeff}x+\\var{pconstant})$.

Factorise $\\simplify{{bp1}a+{bp2}}$

[[0]] $\\large($ [[1]] $\\large)$

Note: Choose the best common factor to minimise the total number of negative signs.

We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

The common factor of $\\simplify{{bp1}a+{bp2}}$ that we should pull out is $\\var{cf}$. This is because:

$\\var{-cf}$ is the largest common factor and
by pulling the negative out the front as well, we avoid having two negative terms in the bracket.

Once we remove that factor from each term in $\\simplify{{bp1}a+{bp2}}$ we are left with $\\var{bx}a+\\var{bc}$.

That means $\\simplify{{bp1}a+{bp2}}$ is $\\var{cf} = \\var{cf}(\\var{bx}a+\\var{bc})$.

Factorise $\\simplify{{ct1}x+{ct2}y+{ct3}}$

[[0]] $\\large($ [[1]] $\\large)$

We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

The (largest) common factor of $\\simplify{{ct1}x+{ct2}y+{ct3}}$ is $\\var{cmult}$.

Once we remove that factor from each term in $\\simplify{{ct1}x+{ct2}y+{ct3}}$ we are left with $\\simplify{{cx}x+{cy}y+{cc}}$.

That means $\\simplify{{ct1}x+{ct2}y+{ct3}} = \\simplify{{cmult}({ct1}x+{ct2}y+{ct3})}$.

Make sure you read instructions on how to enter solutions. If unsure, check what you have entered is the same as what you mean before clicking \"submit\"

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