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{FirstName[0][0]} has written $\\displaystyle{\\simplify{({a}x+{b})/({c}y+{d})}}$ in the equivalent form $\\displaystyle{\\simplify{({ar}x+{br})/({cr}y+{dr})}}$.

What has {FirstName[0][0]} done to the first fraction in order to get the second? {FirstName[0][1]} has divided the top and bottom by [[0]] .

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The two terms in the numerator have a common factor of $\\var{amult}$. The two terms in the denominator also have a common factor of $\\var{amult}$. {FirstName[0][1]} has divided the numerator and denominator by the common factor of $\\var{amult}$ to get the equivalent fraction.

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{FirstName[1][0]} has written $\\displaystyle\\frac{\\simplify{((x+{ar})/({f}x))}}{\\simplify{({br}y+{dr})}}$ in the equivalent form $\\displaystyle{\\simplify{(x+{ar})/({f*br}x*y+{f*dr}x)}}$.

What has {FirstName[1][0]} done to the first fraction in order to get the second? {FirstName[1][1]} has multiplied the top and bottom by [[0]] .

The numerator and denominator have both become $\\var{f}x$ times larger. {FirstName[1][1]} has multiplied the numerator and denominator by $\\var{f}x$ to get an equivalent fraction.

Which of the following are equivalent to $\\displaystyle{\\simplify{({g}x+{h})/(x+{h})}}$?

When you have a fraction, $\\displaystyle{\\simplify{({g}x+{h})/(x+{h})}}$, it is equal to the sum of each term in the numerator being divided by the entire denominator. That is,

\\[\\displaystyle{\\simplify{({g}x+{h})/(x+{h})}}=\\simplify[!collectnumbers]{({g}x)/(x+{h})+({h})/(x+{h})}\\]

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$\\var{g}$

", "

$\\simplify{{g-1}x+{h}}$

", "

$\\displaystyle\\simplify[!collectnumbers]{({g}x)/(x+{h})+({h})/(x+{h})}$

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$\\displaystyle{\\simplify{({j}x+{k})/({l}y)}}$ is equal to:

When you have a fraction, $\\displaystyle{\\simplify{({j}x+{k})/({l}y)}}$, it is equal to the sum of each term in the numerator being divided by the entire denominator. That is,

\\begin{align}\\displaystyle\\simplify{({j}x+{k})/({l}y)}&=\\simplify[!simplifyfractions]{{j}x/({l}y)+{k}/({l}y)}\\\\&=\\simplify{x/(2y)+{k}/({l}y)}\\end{align}

Note: In the last line we cancelled a common factor of $\\var{j}$ in the fraction $\\simplify[!simplifyfractions]{{j}x/({l}y)}$ to get the equivalent fraction $\\simplify{x/(2y)}$.

$\\displaystyle{\\simplify{(x+{k})/(2y)}}$

", "

$\\displaystyle{\\simplify{x/(2y)+{k}}}$

", "

$\\displaystyle{\\simplify{x/(2y)+{k}/({l}y)}}$

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Please fill in the gap to simplify the fraction on the left.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$x+\\var{m}$	=	[[0]]
$(x+\\var{m})(x+\\var{n})$		$x+\\var{n}$

These are equivalent fractions so the same number that multiplied/divided the denominator must also multiply/divide the numerator. Either of the following approaches could be taken:

Notice in the first fraction there is a common factor of $\\simplify{x+{m}}$. To simplify the fraction we would cancel that common factor by dividing the numerator and denominator by the common factor. This would leave only a $1$ in the numerator and $\\simplify{x+{n}}$ in the denominator.
Compare the two denominators, what has happened? We have divided the first denominator by $(x+\\var{m})$ to get the second denominator. The same must be done to the numerator, but something divided by itself is $1$.

The expression $\\displaystyle{\\frac{\\var{p}z}{\\frac{\\var{p}}{\\var{p}z}}}$ can be simplified to [[0]] .

Whenever we have such a 'fraction on a fraction' we want to rewrite the fraction to it is just one fraction. We can do this by multiplying the numerator and denominator by the denominator of the smaller/inner fraction.

In our case, the denominator of the smaller/fraction is $\\var{p}z$:

$\\begin{align}\\displaystyle \\frac{\\var{p}z}{\\frac{\\var{p}}{\\var{p}z}}&=\\frac{\\var{p}z\\times \\var{p}z}{\\frac{\\var{p}}{\\var{p}z}\\times\\var{p}z}\\\\&=\\frac{\\var{p}^2z^2}{\\var{p}}\\\\&=\\var{p}z^2.\\end{align}$

Alternatively, we can think of this as division by a fraction, and we can deal with that using multiplication by the reciprocal:

$\\begin{align}
\\displaystyle \\frac{\\var{p}z}{\\frac{\\var{p}}{\\var{p}z}}
&=\\var{p}z\\div{\\frac{\\var{p}}{\\var{p}z}}\\\\
&= \\var{p}z\\times{\\frac{\\var{p}z}{\\var{p}}}\\\\
&=\\frac{\\var{p}^2z^2}{\\var{p}}\\\\
&=\\var{p}z^2.\\end{align}$

Please fill in the gaps to simplify the fraction on the left.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{{q}x^2y+{r}xy+{s}xy^2}$	=	[[0]]
$\\var{t}xy$		[[1]]

By looking for common factors, we see that $\\var{common}xy$ is a factor of every term (in the numerator and the denominator), no larger term is common, we call $\\var{common}xy$ the highest common factor. We divide the numerator and the denominator by the highest common factor to get the simplified fraction.

Sometimes it is easier to factorise to find the common factor first and then cancel:

$\\begin{align}\\simplify{({q}x^2y+{r}x*y+{s}x*y^2)/({t}x*y)}&=\\dfrac{\\simplify{{common}x*y({list[0]}x+{list[1]}+{list[2]}y)}}{\\simplify{{common}x*y({list[3]})}}\\\\
&=\\dfrac{\\simplify{{list[0]}x+{list[1]}+{list[2]}y}}{\\simplify{{list[3]}}}.\\end{align}$

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