// Numbas version: exam_results_page_options {"name": " Algebra VI: Solving Linear Equation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "b", "c", "ans1", "d", "f", "g", "ans2", "h", "j", "k", "ans3", "l", "m", "n", "ans4", "p", "q", "r", "ans5", "s", "t", "u", "ans6", "v"], "name": " Algebra VI: Solving Linear Equation", "tags": [], "preamble": {"css": "", "js": ""}, "advice": "

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

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Dividing both sides by $\\var{a}$ gives us $x= \\var{ans1}$

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

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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.

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Either way we should end up with $y= \\dfrac{\\simplify{{d- g}}}{\\var{f}}= \\var{ans2}$

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c)

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$\\displaystyle{\\frac{z}{\\var{h}}}-\\var{j}$$=$$\\var{k}$ 
 
 
$\\displaystyle{\\frac{z}{\\var{h}}}$$=$$\\var{k+j}$
 
 
$z$$=$$\\var{ans3}$
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d)

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$\\displaystyle{\\frac{a-\\var{l}}{\\var{m}}}$$=$$\\var{n}$ 
 
$a-\\var{l}$$=$$\\var{n*m}$
 
 
$a$$=$$\\var{ans4}$
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e)

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$\\var{p}$$=$$\\var{q}(\\var{r}+b)$ 
 
 
$\\displaystyle{\\simplify{{p}/{q}}}$$=$$\\var{r}+b$
 
 
$\\displaystyle{\\simplify{{p-r*q}/{q}}}$$=$$b$
$\\var{ans5}$$=$$b$
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f)

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

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$=$\n

$ \\var{ans6}$

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For more help, check this video- 

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 $\\var{a}x+\\var{b}=\\var{c}$

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$x=$ [[0]].

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

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$y=$ [[0]].

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

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$z=$ [[0]]

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

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$a=$ [[0]]

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

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$b=$ [[0]]

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For equations of this type, it would be wise to divide the coefficient of the brackets on both sides and then rearrange as the subject variable $b$ is within the brackets.

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For example,

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Take the equation:    $5=2(1+b)$

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Both sides are divided by the coefficient of the brackets:    $\\frac{5}{2}=1+b$

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Thequation is then rearranged in a typical manner to make $b$ the subject and is simplified:    $b=\\frac{5}{2}-1=\\frac{3}{2}$

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

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$w=$ [[0]]

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Solve the following linear equations using basic BIDMAS calculations ( Answer in two decimal places only)

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For help on solving linear equations, check this video- https://www.youtube.com/watch?v=8j3kRdiRjPA

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

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