Question
Evaluate: $\frac{11}{-12}+\frac{3}{-8}+\frac{1}{4}$
✓
Answer
$\frac{11}{-12}+\frac{3}{-8}+\frac{1}{4}$
$\frac{11}{-12}=\frac{11\times(-1)}{-12\times(-1)}=\frac{-11}{12}$
$\frac{3}{8}=\frac{3\times(-1)}{-8\times(-1)}=\frac{-3}{8}$
Now $LCM of 12,8,4 = 24$
$\begin{array}{c|c}2&12,8,4\\\hline2 &6,4,2\\\hline&3,2,1\end{array}$
$\text{LCM}=2\times2\times2\times3=24$
$\therefore\frac{-11}{12}=\frac{-11\times2}{12\times2}=\frac{-22}{24}$
$\frac{-3}{8}=\frac{-3\times3}{8\times3}=\frac{-9}{24}$
$\frac{1}{4}=\frac{1\times6}{4\times6}=\frac{6}{24}$
$\therefore\frac{-11}{12}+\frac{-3}{8}+\frac{1}{4}$
$=\frac{-22}{24}+\frac{-9}{24}+\frac{6}{24}$
$=\frac{-22+(-9)+6}{24}$
$=\frac{-22-9+6}{24}=\frac{-31+6}{24}$
$=\frac{-25}{24}$
$\frac{11}{-12}=\frac{11\times(-1)}{-12\times(-1)}=\frac{-11}{12}$
$\frac{3}{8}=\frac{3\times(-1)}{-8\times(-1)}=\frac{-3}{8}$
Now $LCM of 12,8,4 = 24$
$\begin{array}{c|c}2&12,8,4\\\hline2 &6,4,2\\\hline&3,2,1\end{array}$
$\text{LCM}=2\times2\times2\times3=24$
$\therefore\frac{-11}{12}=\frac{-11\times2}{12\times2}=\frac{-22}{24}$
$\frac{-3}{8}=\frac{-3\times3}{8\times3}=\frac{-9}{24}$
$\frac{1}{4}=\frac{1\times6}{4\times6}=\frac{6}{24}$
$\therefore\frac{-11}{12}+\frac{-3}{8}+\frac{1}{4}$
$=\frac{-22}{24}+\frac{-9}{24}+\frac{6}{24}$
$=\frac{-22+(-9)+6}{24}$
$=\frac{-22-9+6}{24}=\frac{-31+6}{24}$
$=\frac{-25}{24}$
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