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Subtract the sum of \( \frac{-36}{11} \) and \( \frac{49}{22} \) from the \( \operatorname{sum} \) of \( \frac{33}{8} \) and \( \frac{-19}{4} \).
Given:
Given fractions are $\frac{-36}{11}$ and $\frac{49}{22}$; $\frac{33}{8}$ and $\frac{-19}{4}$.
To do:
Here we have to subtract the sum of $\frac{-36}{11}$ and $\frac{49}{22}$ from the sum of $\frac{33}{8}$ and $\frac{-19}{4}$.
Solution:
Sum of $\frac{-36}{11}$ and $\frac{49}{22}$:
$\frac{-36}{11} \ +\ \frac{49}{22}$
$=\ \frac{2(-36) \ +\ 49}{22}$ (LCM of 11 and 22 is 22)
$=\ \frac{-72\ +\ 49}{22}$
$=\ \mathbf{\frac{-23}{22}}$
Sum of $\frac{33}{8}$ and $\frac{-19}{4}$:
$\frac{33}{8} \ +\ \frac{-19}{4}$
$=\ \frac{33 \ +\ 2(-19)}{8}$ (LCM of 4 and 8 is 8)
$=\ \frac{33\ +\ -38}{8}$
$=\ \mathbf{\frac{-5}{8}}$
Difference of $\frac{-5}{8}$ and $\frac{-23}{22}$:
$\frac{-5}{8}-\frac{-23}{22}$
$=\frac{11(-5)-4(-23)}{88}$ (LCM of 8 and 22 is 88)
$=\frac{-55+92}{88}$
$=\frac{37}{88}$
The answer is $\frac{37}{88}$.