# Simplifying long fractions ((8+\frac{3}{4})+(3\frac{2}{3}))/((4+\frac{2}{5)}-(1\frac{7}{8})) The correct answer is (4+\frac{278}{303})

Simplifying long fractions
$$\displaystyle\frac{{{\left({8}+{\frac{{{3}}}{{{4}}}}\right)}+{\left({3}{\frac{{{2}}}{{{3}}}}\right)}}}{{{\left({4}+{\frac{{{2}}}{{{5}}}}\right\rbrace}-{\left({1}{\frac{{{7}}}{{{8}}}}\right)}}}$$
The correct answer is $$\displaystyle{\left({4}+{\frac{{{278}}}{{{303}}}}\right)}$$

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peterpan7117i
Combine the numerator into an improper fraction:
$$\displaystyle{8}{\frac{{{3}}}{{{4}}}}+{3}{\frac{{{2}}}{{{3}}}}={8}+{3}+{\frac{{{3}}}{{{4}}}}+{\frac{{{2}}}{{{3}}}}$$
$$\displaystyle={\frac{{{132}}}{{{12}}}}+{\frac{{{9}}}{{{12}}}}+{\frac{{{8}}}{{{12}}}}$$
$$\displaystyle={\frac{{{149}}}{{{12}}}}$$
Do the same for the denominator:
$$\displaystyle{4}+{\frac{{{2}}}{{{5}}}}-{1}{\frac{{{7}}}{{{8}}}}={4}-{1}+{\frac{{{2}}}{{{5}}}}-{\frac{{{7}}}{{{8}}}}$$
$$\displaystyle={\frac{{{120}}}{{{30}}}}+{\frac{{{16}}}{{{40}}}}-{\frac{{{35}}}{{{40}}}}$$
$$\displaystyle={\frac{{{101}}}{{{40}}}}$$
Then divide both:
$$\displaystyle{\frac{{{\frac{{{149}}}{{{12}}}}}}{{{\frac{{{101}}}{{{40}}}}}}}={\frac{{{149}}}{{{101}}}}\cdot{\frac{{{40}}}{{{12}}}}$$
$$\displaystyle={\frac{{{149}}}{{{101}}}}\cdot{\frac{{{10}}}{{{3}}}}$$
$$\displaystyle{\frac{{{1490}}}{{{303}}}}$$
$$\displaystyle={4}+{\frac{{{278}}}{{{303}}}}$$
###### Not exactly what you’re looking for?
Beverly Smith
Just simplify it piece by piece.
First, you have
$$\displaystyle{8}+{\frac{{{3}}}{{{4}}}}+{3}{\frac{{{2}}}{{{3}}}}$$
You can simplify this into one fraction.
Then, you simplify
$$\displaystyle{4}{\frac{{{2}}}{{{5}}}}-{1}{\frac{{{7}}}{{{8}}}}$$
into one fraction.
Then use the rule
$$\displaystyle{\frac{{{\frac{{{a}}}{{{b}}}}}}{{{\frac{{{c}}}{{{d}}}}}}}={\frac{{{a}{d}}}{{{b}{c}}}}$$
and you have just one fraction.
Vasquez

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