// Numbas version: exam_results_page_options {"name": "Basic algebra practice", "metadata": {"description": "

Extra practice on some basic algebra skills, including solving linear equations. You can try as many times as you like and also generate new versions of the questions for extra practice.

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Simple exercise in collecting terms in different powers of \\(x\\)

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify the following expression by combining \"like\" terms.

", "advice": "

The idea is to collect together and combine any terms that are the same kind of term so:

\n

$\\var{b}$ and $\\var{f}$ are ordinary numbers. We can combine them to get $\\var{b+f}$

\n

We can combine $\\var{a}x$ and $\\var{d}x$ to get $\\var{a+d}x$.

\n

There are also $\\var{c}$ times $x^2$. So our answer is:

\n

$\\simplify{{c}x^2+{a+d}x+{b+f}}$

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$\\simplify[!collectNumbers]{{a}x+{b}+{c}x^2+{d}x+{f}}$

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Solve linear equations with unknowns on one. Including brackets and fractions.

", "licence": "None specified"}, "statement": "

Solve the following equations to find $x$:

", "advice": "

part a)

\n

\\[\\simplify{{a}x+{b} = {c}} \\]

\n

The first step is to isolate all the $x$-terms to one side of the equation.

\n

To do that we will {add} $\\var{abs(b)}$ onto both sides:

\n

\\[\\var{a}x  = \\simplify{{c-b}} \\]

\n

Now we need to divide both sides by the coefficient of $x$, to leave just one $x$.

\n

Dividing both sides by $\\var{a}$,

\n

\\[  x = {\\simplify{({c}-{b})/{a}}} \\]

\n

\n

part b)

\n

 \\[ \\frac{\\simplify{{d}x + {f}}}{\\var{g}} = \\var{h} \\]

The first step is to rearrange by removing the fraction on the left.  To do this we mulitply both sides by $\\var{g}$.

\\[ \\begin{split} \\frac{\\simplify{{d}x + {f}}}{\\var{g}} \\times \\var{g}  &= \\var{h} \\times \\var{g} \\\\\\\\ \\simplify{{d}x + {f}} &= \\var{h*g} \\end{split} \\]

Next we isolate all the $x$-terms onto one side of the equation.

To do that we will {add2} $\\var{abs(f)}$ on both sides:

\\[ \\begin{split} \\var{d}x &= \\simplify[]{{h*g}-{f}} \\\\ &= \\simplify[]{{h*g-f}} \\end{split}\\]

Finally we need to divide both sides by the coefficient of $x$, to leave just one $x$.

\n

Dividing both sides by $\\var{d}$,

\\[x = \\simplify[fractionNumbers]{{(h*g-f)/d}}  \\]

\n

\n

part c)

\n

\\[ \\simplify{{b}({c}x+{g})} = \\var{d} \\]

\n

Even though this looks different, this is quite similar to part b).  We just have a multiplication rather than a division to deal with as the first step.

Rearrange to remove the multiplication on the left.  To do this we divide both sides by $\\var{b}$:

\\[ \\begin{split} \\frac{\\simplify{{b}({c}x+{g})}}{ \\var{b}} &= \\frac{\\var{d}}{\\var{b}} \\\\ \\\\ \\simplify{{c}x+{g}} &= \\simplify[fractionNumbers]{{d/b}} \\end{split} \\]


Next we isolate all the $x$-terms onto one side of the equation.

To do that we will {add3} $\\var{abs(g)}$ on both sides:

\\[ \\begin {split} \\simplify{{c}x + {g}}   \\var{add3sym}   \\var{abs(g)} &= \\simplify[fractionNumbers]{{d/b}} \\var{add3sym} \\var{abs(g)} \\\\ \\var{c}x &= \\simplify[fractionNumbers]{{d/b-g}} \\end{split} \\]

Finally we need to divide both sides by the coefficient of $x$, to leave just one $x$

\n

Dividing both sides by $\\var{c}$:

\\[ x =  \\simplify[fractionNumbers]{{(d/b-g)/c}} \\]

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

\n

\n

$x=$ [[0]]

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$\\dfrac{\\simplify{{d}x + {f}}}{\\var{g}} = \\var{h}$

\n

\n

$x=$ [[0]]

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$\\simplify{{b}({c}x+{g})} = \\var{d}$

\n

\n

$x=$ [[0]]

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Solve linear equations with unkowns on both sides. Including brackets and fractions.

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

Part a)

\n

Given $\\simplify{{l}({m}w-{n}) = {p}w+{q}}$, 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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\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{{l}({m}w-{n})}$$=$$\\simplify{{p}w+{q}}$ 
 
$\\simplify{{l*m}w+{n*l}}$$=$$\\simplify{{p}w+{q}}$
 
$\\simplify[!cancelTerms,unitFactor]{{l*m}w-{n*l}-{p}w}$$=$$\\simplify[!cancelTerms,unitFactor]{{p}w+{q}-{p}w}$
 
$\\simplify{{l*m-p}w-{n*l}}$$=$$\\var{q}$
 
$\\var{l*m-p}w-\\var{n*l}+\\var{n*l}$$=$$\\var{q}+\\var{n*l}$
 
$\\var{l*m-p}w$$=$$\\var{q+n*l}$
 
$\\displaystyle{\\frac{\\var{l*m-p}w}{\\var{l*m-p}}}$$=$$\\displaystyle{\\frac{\\var{q+n*l}}{\\var{l*m-p}}}$
 
$w$$=$$\\displaystyle{\\simplify{{q+n*l}/{l*m-p}}} = \\var{precround(ansA,1)} \\text{ to 1 dp}$
\n

\n

Part b)

\n

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\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$$=$$\\simplify[unitFactor]{{g}y+{-g*f}}$
 
$\\simplify[!cancelTerms,unitFactor]{{d}y+{-g}y}$ $=$$\\simplify[!cancelTerms,unitFactor]{{g}y+{-g*f}+{-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}}}= \\var{precround(ansB,1)}\\text{ to 1 dp}$
\n

part c)

\n

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\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})}}= \\var{precround(ansD,1)} \\text{ to 1 dp}$(divide by the coefficient of $x$)
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"group": "Ungrouped variables", "definition": "top/sumdeno", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "100"}, "ungrouped_variables": ["a", "b", "c", "ans1", "d", "f", "g", "ans2", "h", "j", "k", "ans3", "l", "m", "n", "p", "q", "ans4", "ansA", "ansB", "ansC", "ansD"], "variable_groups": [{"name": "e", "variables": ["primes", "denom1", "denom2", "add", "right", "eans", "r1", "a2", "r12", "top", "sumdeno"]}], "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": "

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

\n

$w=$ [[0]]

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

\n

$y=$ [[0]].

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

\n

$x=$ [[0]]

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