Simplifying long fractions 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.
 A: 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:
$$\begin{align}\frac{\frac{149}{12}}{\frac{101}{40}} &= \frac{149}{101}\cdot\frac{40}{12}\\
&= \frac{149}{101}\cdot\frac{10}{3}\\
&=\frac{1490}{303}\\
&=4 + \frac{278}{303}\end{align}$$
A: 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.
A: Expand $\dfrac{8+\frac{3}{4}+3+\frac{2}{3}}{4+\frac{2}{5}-1-\frac{7}{8}}$ by $120$ to get
$\dfrac{11\cdot120+3\cdot30+2\cdot40}{3\cdot120+2\cdot24-7\cdot15}=\dfrac{1490}{303}.$
