Question
Find $\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\left(\frac{5}{22}\right)$
✓
Answer
We have, $\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\left(\frac{5}{22}\right)$
$=\frac{198}{462}+\left(\frac{-252}{462}\right)+\left(\frac{-176}{462}\right)+\left(\frac{105}{462}\right)$[$\because$ $462$ is the $LCM$ of $7, 11, 21$ and $22$]
$=\frac{198-252-176+105}{462}$
$=\frac{-125}{462}$
Alternatively: We can also solve it as.
$\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\frac{5}{22}$
$=\left[\frac{3}{7}+\left(\frac{-8}{21}\right)\right]+\left[\frac{-6}{11}+\frac{5}{22}\right]$
$=\left[\frac{9+(-8)}{21}\right]+\left[\frac{-12+5}{22}\right]$[$\because$$LCM$ of $7$ and $21$ is $21$; $LCM$ of $11$ and $22$ is $22$]
$=\frac{1}{21}+\left(\frac{-7}{22}\right)$
$=\frac{22-147}{462}$
$=\frac{-125}{462}$
$=\frac{198}{462}+\left(\frac{-252}{462}\right)+\left(\frac{-176}{462}\right)+\left(\frac{105}{462}\right)$[$\because$ $462$ is the $LCM$ of $7, 11, 21$ and $22$]
$=\frac{198-252-176+105}{462}$
$=\frac{-125}{462}$
Alternatively: We can also solve it as.
$\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\frac{5}{22}$
$=\left[\frac{3}{7}+\left(\frac{-8}{21}\right)\right]+\left[\frac{-6}{11}+\frac{5}{22}\right]$
$=\left[\frac{9+(-8)}{21}\right]+\left[\frac{-12+5}{22}\right]$[$\because$$LCM$ of $7$ and $21$ is $21$; $LCM$ of $11$ and $22$ is $22$]
$=\frac{1}{21}+\left(\frac{-7}{22}\right)$
$=\frac{22-147}{462}$
$=\frac{-125}{462}$
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