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

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

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If we multiply or divide the top and bottom of a fraction by a number (not zero) we get an equivalent fraction. We say equivalent because they represent the same amount of the whole.

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Recall $\\frac{3}{6}$ is equivalent to $\\frac{1}{2}$, notice we can divide the top and bottom of $\\frac{3}{6}$ by 3 to get $\\frac{1}{2}$. Similarly, if we divide the top and bottom of $\\frac{10x-50y}{20a+10}$ by $10$ we would get the equivalent fraction $\\frac{x-5y}{2a+1}$. 

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When simplifying fractions we try to get the fraction so there is no common factor on the top and the bottom (if there was we would divide by it). It is important that this common factor is common to all the terms and not just a couple, otherwise dividing by it would leave you with a fraction on a fraction.

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

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

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If we multiply or divide the top and bottom of a fraction by a number (not zero) we get an equivalent fraction. We say equivalent because they represent the same amount of the whole.

\n

\n

Recall $\\frac{1}{2}$ is equivalent to $\\frac{3}{6}$, notice we can multiply the top and bottom of $\\frac{1}{2}$ by 3 to get $\\frac{3}{6}$. Similarly, we can multiply the top and bottom of $\\frac{\\frac{x-3}{2x}}{x+1}$ by $2x$ to get the equivalent fraction $\\frac{x-3}{2x^2+2x}$. 

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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 by the denominator of the smaller/inner fraction.

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

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When you have a fraction, say $\\displaystyle{\\frac{1+2a+3b}{5}}$, it represents the result of dividing everything on top by everything on the bottom, that is 

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\\[\\frac{1+2a+3b}{5+x}=\\frac{1}{5+x}+\\frac{2a}{5+x}+\\frac{3b}{5+x}\\]

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$\\var{g}$

", "

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

", "

$\\displaystyle{\\simplify{({g}x)/(x+{h})}+\\simplify{({h})/(x+{h})}}$

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

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When you have a fraction, say $\\displaystyle{\\frac{1+2a+3b}{5}}$, it represents the result of dividing everything on top by everything on the bottom, that is 

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\\[\\frac{1+2a+3b}{5+x}=\\frac{1}{5+x}+\\frac{2a}{5+x}+\\frac{3b}{5+x}\\]

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

", "

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

", "

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

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\n

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}$
\n
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These are equivalent fractions so the same number that multiplied/divided the denominator must also multiply/divide the numerator.

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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 $1$.

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\n

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

\n
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If we multiply or divide the top and bottom of a fraction by a number (not zero) we get an equivalent fraction. We say equivalent because they represent the same amount of the whole.

\n

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Recall $\\frac{1}{2}$ is equivalent to $\\frac{3}{6}$, notice we can multiply the top and bottom of $\\frac{1}{2}$ by 3 to get $\\frac{3}{6}$. Similarly, we can multiply the top and bottom of $\\frac{\\frac{x-3}{2x}}{x+1}$ by $2x$ to get the equivalent fraction $\\frac{x-3}{2x^2+2x}$. 

\n

\n

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 by the denominator of the smaller/inner fraction.

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\n

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
$\\simplify{{q}x^2y+{r}x*y+{s}x*y^2}$=[[0]]
$\\var{t}xy$[[1]]
\n
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By looking for common factors, we see that $\\var{common}xy$ is a factor of every term (in the numerator and the denominator), there is also no larger term that 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.

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