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 $\frac{37}{88}$.

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Updated on: 10-Oct-2022

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