I need someone to help me understand how to do these problems.
Your images are hard to read. You might want to try to find another method to present your work. This is what I think you have. Please confirm and others will be glad to assist.
a) $\displaystyle \frac{-2(18-2)}{4(5-6)}-\frac{3(27-37)}{5(7-4)}$
b) $\displaystyle \frac{\frac{3}{4}-7}{\frac{1}{2}-\frac{4}{5}}$
c) $\displaystyle \frac{\frac{3}{8}}{\frac{1}{4}-\frac{7}{16}}$
d) $\displaystyle 4 \frac{3}{5}+7 \frac{1}{10}$
e) $\displaystyle 9\left(\frac{15}{81}\right)\left(\frac{27}{15}\rig ht)$
your pics are kinda blurry, so make sure i got the problem right.
Get like denominators on both the top and bottom of the big fraction.$\displaystyle \frac{\frac{3}{4} - 7 }{\frac{1}{2} - \frac{4}{5}}$
$\displaystyle \frac{\frac{3}{4} - \frac{28}{4}}{\frac{5}{10} - \frac{8}{10}}$
Simplify: $\displaystyle \frac{\frac{-25}{4}}{\frac{-3}{10}}$
Multiply by the reciprocal: $\displaystyle \frac{-25}{4}\cdot\frac{-10}{3}$
$\displaystyle = \frac{250}{12}$
$\displaystyle = \frac{125}{6}$
C is very similar.
On A, do what's in parenthesis first, get a like denominator and simplify.
For: $\displaystyle 4 \frac{3}{5}+7 \frac{1}{10}$, get improper fractions, then a like denominator and finally simplify.
For: $\displaystyle 9\left(\frac{15}{81}\right)\left(\frac{27}{15}\rig ht)$
Multiply straight across; numerator by numerator; denominator by denominator.
You can think of it as: $\displaystyle \bigg(\frac{9}{1}\bigg)\left(\frac{15}{81}\right)\ left(\frac{27}{15}\right)$
Look for cross canceling. The 15 is an obvious one.
Then simplify as much as possible.
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