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| $\\var{a}x+\\var{b}$ | \n$=$ | \n$\\var{c}x$ | \n
| \n | \n | \n |
| $\\var{a}x+\\var{b}-\\var{a}x$ | \n$=$ | \n$\\var{c}x-\\var{a}x$ | \n
| \n | \n | \n |
| $\\var{b}$ | \n$=$ | \n$\\var{c-a}x$ | \n
| \n | \n | \n |
| $\\displaystyle{\\frac{\\var{b}}{\\var{c-a}}}$ | \n$=$ | \n$\\displaystyle{\\frac{\\var{c-a}x}{\\var{c-a}}}$ | \n
| \n | \n | \n |
| $\\displaystyle{\\simplify{{b}/{c-a}}}$ | \n$=$ | \n$x$ | \n
| \n | \n | \n |
| $x$ | \n$=$ | \n$\\displaystyle{\\simplify{{b}/{c-a}}}$ | \n
There is often more than one way to solve an equation, one strategy used above in the first step was to get all the $x$'s one the side with the most $x$'s, that way you end up with a postive number of $x$'s. This is not necessary, we could have put the all the $x$'s on the left hand side but notice in this question we then would have had to move the $\\var{b}$ on to the right hand side, so it would have required more work, but nevertheless that method would result in the same result for $x$.
", "marks": 0, "showCorrectAnswer": true}], "variableReplacements": [], "customName": "", "type": "gapfill", "customMarkingAlgorithm": "", "useCustomName": false, "variableReplacementStrategy": "originalfirst", "scripts": {}, "gaps": [{"customMarkingAlgorithm": "", "correctAnswerStyle": "plain", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "customName": "", "maxValue": "{-b}/{a-c}", "type": "numberentry", "unitTests": [], "correctAnswerFraction": true, "useCustomName": false, "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "scripts": {}, "variableReplacements": [], "minValue": "{-b}/{a-c}", "showFractionHint": true, "allowFractions": true, "marks": 1, "showCorrectAnswer": true}], "prompt": "We can solve $\\var{a}x+\\var{b}=\\var{c}x$ by collecting like terms and rearranging for $x$. This gives $x=$ [[0]].
", "marks": 0, "showCorrectAnswer": true, "unitTests": []}, {"sortAnswers": false, "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "stepsPenalty": "1", "steps": [{"extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "unitTests": [], "variableReplacements": [], "customName": "", "type": "information", "customMarkingAlgorithm": "", "useCustomName": false, "variableReplacementStrategy": "originalfirst", "scripts": {}, "prompt": "Given $\\var{l}(\\var{m}w-\\var{n})=\\var{p}w+\\var{q}$, we can expand the brackets, get all the $w$'s on the left hand side and all the numbers on the right hand side, and then divide both sides by the coefficient of $w$ to get $w$ by itself.
\n\n| $\\var{l}(\\var{m}w-\\var{n})$ | \n$=$ | \n$\\var{p}w+\\var{q}$ | \n
| \n | \n | \n |
| $\\var{l*m}w-\\var{n*l}$ | \n$=$ | \n$\\var{p}w+\\var{q}$ | \n
| \n | \n | \n |
| $\\var{l*m}w-\\var{n*l}-\\var{p}w$ | \n$=$ | \n$\\var{p}w+\\var{q}-\\var{p}w$ | \n
| \n | \n | \n |
| $\\var{l*m-p}w-\\var{n*l}$ | \n$=$ | \n$\\var{q}$ | \n
| \n | \n | \n |
| $\\var{l*m-p}w-\\var{n*l}+\\var{n*l}$ | \n$=$ | \n$\\var{q}+\\var{n*l}$ | \n
| \n | \n | \n |
| $\\var{l*m-p}w$ | \n$=$ | \n$\\var{q+n*l}$ | \n
| \n | \n | \n |
| $\\displaystyle{\\frac{\\var{l*m-p}w}{\\var{l*m-p}}}$ | \n$=$ | \n$\\displaystyle{\\frac{\\var{q+n*l}}{\\var{l*m-p}}}$ | \n
| \n | \n | \n |
| $w$ | \n$=$ | \n$\\displaystyle{\\simplify{{q+n*l}/{l*m-p}}}$ | \n
Solve $\\var{l}(\\var{m}w-\\var{n})=\\var{p}w+\\var{q}$ for $w$.
\n$w=$ [[0]]
", "marks": 0, "showCorrectAnswer": true, "unitTests": []}, {"sortAnswers": false, "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "stepsPenalty": "1", "steps": [{"extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "unitTests": [], "variableReplacements": [], "customName": "", "type": "information", "customMarkingAlgorithm": "", "useCustomName": false, "variableReplacementStrategy": "originalfirst", "scripts": {}, "prompt": "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 $y$'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| $\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}$ | \n$=$ | \n$\\var{g}$ | \n
| \n | \n | \n |
| $\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}\\times(y-\\var{f})$ | \n$=$ | \n$\\var{g}\\times (y-\\var{f})$ | \n
| \n | \n | \n |
| $\\var{d}y$ | \n$=$ | \n$\\var{g}y+\\var{-g*f}$ | \n
| \n | \n | \n |
| $\\var{d}y+\\var{-g}y$ | \n$=$ | \n$\\var{g}y+\\var{-g*f}+\\var{-g}y$ | \n
| \n | \n | \n |
| $\\var{d-g}y$ | \n$=$ | \n$\\var{-g*f}$ | \n
| \n | \n | \n |
| $\\displaystyle{\\frac{\\var{d-g}y}{\\var{d-g}}}$ | \n$=$ | \n$\\displaystyle{\\frac{\\var{-g*f}}{\\var{d-g}}}$ | \n
| \n | \n | \n |
| $y$ | \n$=$ | \n$\\displaystyle{\\simplify{{-g*f}/{d-g}}}$ | \n
Given $\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}=\\var{g}$, $y=$ [[0]].
", "marks": 0, "showCorrectAnswer": true, "unitTests": []}, {"sortAnswers": false, "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "stepsPenalty": "1", "steps": [{"extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "unitTests": [], "variableReplacements": [], "customName": "", "type": "information", "customMarkingAlgorithm": "", "useCustomName": false, "variableReplacementStrategy": "originalfirst", "scripts": {}, "prompt": "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 $z$'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| $\\displaystyle{\\frac{z+\\var{h}}{z+\\var{j}}}$ | \n$=$ | \n$\\var{k}$ | \n
| \n | \n | \n |
| $\\displaystyle{\\frac{z+\\var{h}}{z+\\var{j}}}\\times(z+\\var{j})$ | \n$=$ | \n$\\var{k}\\times (z+\\var{j})$ | \n
| \n | \n | \n |
| $z+\\var{h}$ | \n$=$ | \n$\\var{k}z+\\var{k*j}$ | \n
| \n | \n | \n |
| $z+\\var{h}-\\var{k}z$ | \n$=$ | \n$\\var{k}z+\\var{k*j}-\\var{k}z$ | \n
| \n | \n | \n |
| $\\var{1-k}z+\\var{h}$ | \n$=$ | \n$\\var{k*j}$ | \n
| \n | \n | \n |
| $\\var{1-k}z+\\var{h}-\\var{h}$ | \n$=$ | \n$\\var{k*j}-\\var{h}$ | \n
| \n | \n | \n |
| $\\var{1-k}z$ | \n$=$ | \n$\\var{k*j-h}$ | \n
| \n | \n | \n |
| $\\displaystyle{\\frac{\\var{1-k}z}{\\var{1-k}}}$ | \n$=$ | \n$\\displaystyle{\\frac{\\var{k*j-h}}{\\var{1-k}}}$ | \n
| \n | \n | \n |
| $z$ | \n$=$ | \n$\\displaystyle{\\simplify{({k*j-h})/({1-k})}}$ | \n
Solve $\\displaystyle{\\frac{z+\\var{h}}{z+\\var{j}}}=\\var{k}$ for $z$.
\n$z=$ [[0]]
", "marks": 0, "showCorrectAnswer": true, "unitTests": []}, {"sortAnswers": false, "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "stepsPenalty": "1", "steps": [{"extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "unitTests": [], "variableReplacements": [], "customName": "", "type": "information", "customMarkingAlgorithm": "", "useCustomName": false, "variableReplacementStrategy": "originalfirst", "scripts": {}, "prompt": "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 $x$'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| $\\displaystyle{\\frac{x+\\var{add}}{\\var{denom1}}+\\frac{x}{\\var{denom2}}}$ | \n$=$ | \n$\\var{right}$ | \n\n |
| \n | \n | \n | \n |
| $\\displaystyle{\\left(\\frac{x+\\var{add}}{\\var{denom1}}\\right)\\times\\var{denom1}+\\left(\\frac{x}{\\var{denom2}}\\right)\\times\\var{denom1}}$ | \n$=$ | \n$\\var{right}\\times \\var{denom1}$ | \n(multiply all terms by $\\var{denom1}$) | \n
| \n | \n | \n | \n |
| $\\displaystyle{x+\\var{add}+\\frac{\\var{denom1}x}{\\var{denom2}}}$ | \n$=$ | \n$\\var{r1}$ | \n\n |
| \n | \n | \n | \n |
| $\\displaystyle{(x+\\var{add})\\times\\var{denom2}+\\left(\\frac{\\var{denom1}x}{\\var{denom2}}\\right)\\times\\var{denom2}}$ | \n$=$ | \n$\\var{r1}\\times\\var{denom2}$ | \n(multiply all terms by $\\var{denom2}$) | \n
| \n | \n | \n | \n |
| $\\displaystyle{\\var{denom2}x+\\var{a2}+\\var{denom1}x}$ | \n$=$ | \n$\\var{r12}$ | \n\n |
| \n | \n | \n | \n |
| $\\var{sumdeno}x+\\var{a2}$ | \n$=$ | \n$\\var{r12}$ | \n(collect like terms) | \n
| \n | \n | \n | \n |
| $\\var{sumdeno}x$ | \n$=$ | \n$\\var{r12}-\\var{a2}$ | \n(collect like terms) | \n
| \n | \n | \n | \n |
| $\\var{sumdeno}x$ | \n$=$ | \n$\\var{top}$ | \n\n |
| \n | \n | \n | \n |
| $x$ | \n$=$ | \n$\\displaystyle{\\simplify{{top}/({sumdeno})}}$ | \n(divide by the coefficient of $x$) | \n
Solve $\\displaystyle{\\frac{x+\\var{add}}{\\var{denom1}}+\\frac{x}{\\var{denom2}}=\\var{right}}$.
\n$x=$ [[0]]
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| \n | \n | \n |
| $\\var{a}x+\\var{b}-\\var{a}x$ | \n$=$ | \n$\\var{c}x-\\var{a}x$ | \n
| \n | \n | \n |
| $\\var{b}$ | \n$=$ | \n$\\var{c-a}x$ | \n
| \n | \n | \n |
| $\\displaystyle{\\frac{\\var{b}}{\\var{c-a}}}$ | \n$=$ | \n$\\displaystyle{\\frac{\\var{c-a}x}{\\var{c-a}}}$ | \n
| \n | \n | \n |
| $\\displaystyle{\\simplify{{b}/{c-a}}}$ | \n$=$ | \n$x$ | \n
| \n | \n | \n |
| $x$ | \n$=$ | \n$\\displaystyle{\\simplify{{b}/{c-a}}}$ | \n