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How would I go about simplifying long fractions, such as the likes of this:

$((8+\frac{3}{4}) + (3\frac{2}{3}))$ / $((4+\frac{2}{5}) - (1\frac{7}{8}))$

The correct answer is ($4 + \frac{278}{303}$)

I'm not really sure how to approach this problem,

regards.

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  • $\begingroup$ Welcome to Math StackExchange! How about trying to simplify the inside expressions first and tell us what you get? $\endgroup$ – user21820 Feb 16 '14 at 10:25
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Combine the numerator into an improper fraction: $$\begin{align}8 + \frac{3}{4} + 3\frac{2}{3} &= 8 + 3 + \frac{3}{4} + \frac{2}{3} \\ &= \frac{132}{12} + \frac{9}{12} + \frac{8}{12} \\ &= \frac{149}{12}\end{align}$$

Do the same for the denominator: $$\begin{align}4 + \frac{2}{5} - 1\frac{7}{8} &= 4 - 1 + \frac{2}{5} - \frac{7}{8}\\ &=\frac{120}{30} + \frac{16}{40} - \frac{35}{40} \\ &= \frac{101}{40}\end{align}$$

Then divide both: $$\begin{align}\frac{\frac{149}{12}}{\frac{101}{40}} &= \frac{149}{101}\cdot\frac{40}{12}\\ &= \frac{149}{101}\cdot\frac{10}{3}\\ &=\frac{1490}{303}\\ &=4 + \frac{278}{303}\end{align}$$

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Just simplify it piece by piece.

First, you have $$8+\frac 34 + 3\frac 23$$ You can simplify this into one fraction.

Then, you simplify $$4+\frac25 - 1\frac78$$ into one fraction.

Then use the rule $$\frac{\frac ab}{\frac cd} = \frac{ad}{bc}$$ and you have just one fraction.

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Expand $\dfrac{8+\frac{3}{4}+3+\frac{2}{3}}{4+\frac{2}{5}-1-\frac{7}{8}}$ by $120$ to get $\dfrac{11\cdot120+3\cdot30+2\cdot40}{3\cdot120+2\cdot24-7\cdot15}=\dfrac{1490}{303}.$

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