// Numbas version: exam_results_page_options {"name": "Negatives: subtracting negative numbers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Negatives: subtracting negative numbers", "tags": ["negative numbers", "Negative Numbers", "negatives", "subtraction"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "
Complete the following without the use of a calculator:
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", "stepsPenalty": "1", "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": "Thinking about negative numbers as debt can help you understand these questions.
\nWe can think of $\\var{n[0]}$ as being in debt by \\${-n[0]}. Suppose we then 'take away' another debt of \\${-n[1]}, that is, we pay back \\${-n[1]}. Now we are only in debt actually in front by \\${abs(ans1)}. That is, $\\var{n[0]}-(\\var{n[1]})=\\var{ans1}$.
\nAnother way is to think of the minus sign as going in the reverse direction.
\nSuppose on the number line we are at $\\var{n[0]}$, and now instead of going another $\\var{-n[1]}$ to the left (which we would do if we were doing $\\var{n[0]}+(\\var{n[1]})$) we have to go the opposite way, we have to move $\\var{-n[1]}$ to the right (which is the same as $\\var{n[0]}+\\var{-n[1]}$, and we end up at $\\var{ans1}$.
\nIn general two negative symbols next to each other are the same as a positive symbol.
\n\\[++=+\\]
\n\\[+-=-\\]
\n\\[-+=-\\]
\n\\[--=+\\]
\nAny even number of negative symbols will be the same as a positive symbol. For example,
\n\\[-----=+-=-\\]
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", "stepsPenalty": "1", "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": "This is actually the same process as the last question. The lack of brackets doesn't affect this question.
\nThinking about negative numbers as debt can help you understand these questions.
\nWe can think of $\\var{n[2]}$ as being in debt by \\${-n[2]}. Suppose we then 'take away' another debt of \\${-n[3]}, that is, we pay back \\${-n[3]}. Now we are only in debt actually in front by \\${abs(ans2)}. That is, $\\var{n[2]}-(\\var{n[3]})=\\var{ans2}$.
\nAnother way is to think of the minus sign as going in the reverse direction.
\nSuppose on the number line we are at $\\var{n[2]}$, and now instead of going another $\\var{-n[3]}$ to the left (which we would do if we were doing $\\var{n[2]}+(\\var{n[3]})$) we have to go the opposite way, we have to move $\\var{-n[3]}$ to the right (which is the same as $\\var{n[2]}+\\var{-n[3]}$, and we end up at $\\var{ans2}$.
\nIn general two negative symbols next to each other is the same as a positive symbol.
\n\\[++=+\\]
\n\\[+-=-\\]
\n\\[-+=-\\]
\n\\[--=+\\]
\nAny even number of negative symbols will be the same as a positive symbol. For example,
\n\\[-----=+-=-\\]
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