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\n

Part a)

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Given $\\var{a}x+\\var{b}=\\var{c}$, we can start by subtracting $\\var{b}$ from both sides to get $\\var{a}x = \\simplify{{c-b}}$.

\n

Dividing both sides by $\\var{a}$ gives us $x= \\simplify {{c-b}/{a}}$

\n

\n


Part b)

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We could start by subtracting $\\var{d}$ from both sides to get $-\\var{f}y = \\simplify{{g-d}}$.

\n

Or we could first add $\\var{f}y $ to both sides to get $\\var{d} = \\var{g} + \\var{f}y$. This avoids needing to divide by a negative number.

\n

Either way we should end up with $y= \\dfrac{\\simplify{{d- g}}}{\\var{f}}= \\simplify{{d-g}/{f}}$.

\n

\n

\n

Part c)

\n

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

\n

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

\n

$z=\\var{ans3}$

\n

\n

\n

Part d)

\n

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

\n

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

\n

$a=\\var{ans4}$

\n



\n

\n

Part e)

\n

$\\var{p}$$=$$\\var{q}(\\var{r}+b)$ 

\n

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

\n

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

\n

\n

\n

Part f)

\n

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

\n

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

\n

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

\n

\n


For more help, check this video- 

\n

                                               

\n

\n

", "statement": "

Solve the following equations. In each case give your answer to 2 decimal places.

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$x=$ [[0]].

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

\n

$y=$ [[0]].

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

\n

$z=$ [[0]]

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

\n

$a=$ [[0]]

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

\n

$b=$ [[0]]

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

\n

$w=$ [[0]]

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This exercise will help you solve equations of type ax-b = c.

\n

"}, "variable_groups": [], "functions": {}, "variables": {"a": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "a", "definition": "random(2..12)"}, "h": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "h", "definition": "random(list(2..12)+[20,50,100,200])"}, "g": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "g", "definition": "random(-12..12 except [0])"}, "k": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "k", "definition": "random(-13..-1 except -j)"}, "u": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "u", "definition": "random(-12..12 except 0)"}, "c": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "c", "definition": "random(-12..12 except [0,b])"}, "p": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "p", "definition": "random(-12..12 except[0,q])"}, "n": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "n", "definition": "random(-12..12)"}, "ans2": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "ans2", "definition": "(g-d)/(-f)"}, "t": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "t", "definition": "random([13,11,7,5,3,2] except s)"}, "l": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "l", "definition": "random(2..12)"}, "d": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "d", "definition": "random(1..12 except [f,g])"}, "ans3": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "ans3", "definition": "(k+j)*h"}, "b": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "b", "definition": "random(2..12 except a)"}, "ans4": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "ans4", "definition": "n*m+l"}, "ans5": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "ans5", "definition": "p/q-r"}, "ans1": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "ans1", "definition": "(c-b)/a"}, "s": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "s", "definition": "random([-13,-11,-7,-5,-3,-2,13,11,7,5,3,2])"}, "j": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "j", "definition": "random(2..12 except [h])"}, "q": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "q", "definition": "random(-12..12 except [0,1,-1])"}, "f": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "f", "definition": "random(2..12)"}, "ans6": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "ans6", "definition": "u*t/s"}, "r": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "r", "definition": "random(1..12)"}, "m": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "m", "definition": "random(2..12 except l)"}}, "type": "question"}, {"name": "Solving equations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Hannah Bartholomew", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/530/"}], "parts": [{"variableReplacements": [], "showCorrectAnswer": true, "prompt": "

$\\simplify{{a}(x+{b})={c}x-{d}}$

\n

$x =$ [[0]]

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$\\simplify{{m}(x +{sign1} {f})/{g}+ {sign2} {h}} = \\simplify{{j}x/{k}+ {sign3} {l}}$

\n

$x=$ [[0]]

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Solve the following equations.

\n

Please give your answer as a whole number or as a fraction in the form a/b where a and b are integers.

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f

", "name": "f"}, "h": {"definition": "random(1..10#1)", "group": "Ungrouped variables", "templateType": "randrange", "description": "", "name": "h"}, "c": {"definition": "random(3..10#1)", "group": "Ungrouped variables", "templateType": "randrange", "description": "", "name": "c"}, "l": {"definition": "random(0..6#1)", "group": "Ungrouped variables", "templateType": "randrange", "description": "", "name": "l"}, "sign2": {"definition": "random(1,-1)", "group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "sign2"}, "gcd": {"definition": "gcd(em,g)", "group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "gcd"}, "gcd1": {"definition": "gcd(j,k)", "group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "gcd1"}, "sign3": {"definition": "random(1,-1)", "group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "sign3"}, "b": {"definition": "random(3..10#1)", "group": "Ungrouped variables", "templateType": "randrange", "description": "", "name": "b"}, "g": {"definition": "random(4..9#1)", "group": "Ungrouped variables", "templateType": "randrange", "description": "", "name": "g"}, "sign1": {"definition": "random(1, -1)\n", "group": "Ungrouped variables", "templateType": "anything", "description": "

sign1

", "name": "sign1"}}, "variablesTest": {"condition": "a<>b\nm<>g\nj<>k\nj/k<>m/g\na<>c", "maxRuns": "182"}, "metadata": {"licence": "None specified", "description": ""}, "rulesets": {}, "variable_groups": [{"variables": [], "name": "Unnamed group"}], "type": "question"}, {"name": "Hannah's copy of Solve equations which include a single root (e.g. \\sqrt{x}=blah)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Hannah Bartholomew", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/530/"}, {"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "variable_groups": [{"variables": ["intpower", "intrhs", "intsoln"], "name": "a"}, {"variables": ["bpower", "bnice", "bsoln", "bxcoeff", "bb", "bc"], "name": "b"}], "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

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

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

", "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "marks": 0}, {"variableReplacements": [], "variableReplacementStrategy": "originalfirst", "scripts": {}, "gaps": [{"showpreview": true, "checkingaccuracy": 0.001, "variableReplacementStrategy": "originalfirst", "checkvariablenames": false, "vsetrangepoints": 5, "expectedvariablenames": [], "answersimplification": "basic", "showFeedbackIcon": true, "variableReplacements": [], "marks": 1, "checkingtype": "absdiff", "showCorrectAnswer": true, "vsetrange": [0, 1], "scripts": {}, "type": "jme", "answer": "({dc*ddenom})^({dpower})-{db}"}], "prompt": "

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

\n

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

", "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "marks": 0}], "advice": "

a) 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}$
\n

\n

b) 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}$
\n

\n

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

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

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Please complete the following.

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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

\n

\\[11-25=-(25-11)=-14.\\] 

\n

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 $-1$.

\n

\n

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

\n

\n

a)

\n

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!

\n

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

\n

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

\n

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

\n

\n

b) 

\n

\n

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,

\n

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

\n

So the answers are the same!

\n

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:

\n

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! 

\n

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Equal

", "

Not equal

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Suppose you do a maths question and your answer is 

\n

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

\n

However, your friend has an answer of

\n

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

\n

\n

These answers are...

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Suppose you do a maths question and your answer is:

\n

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

\n

However, your friend has an answer of:

\n

\\[z=\\simplify{(-{a}x^{p}+{sign1}{b}y^{q})/({sign2}{c}x*y+{sign3}{d}x^{r}y^{s})}\\]

\n

\n

Which of the statements below is true?

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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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1

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r

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$\\simplify{{a}x={c-b}}$", "
$\\simplify{{a}x={c+b}}$", "
$\\simplify{x+{b}={c/a}}$", "
$\\simplify{x+{b}={c*a}}$", "
$\\simplify{{a}x={b-c}}$", "
$\\simplify{{a}x={c/b}}$", "
$\\simplify{x+{b}={c-a}}$"], "prompt": "

A class is set the equation $\\simplify{{a}x+{b}={c}}$ to solve.

\n

Which one of the the six equations below is a valid step in rearranging the equation to solve it?

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For each pair of algebraic expressions below, decide if you think they are mathematically equivalent. Check the box if you believe the expressions to be equivalent to each other.

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