Question

Question 94 (i)

Taking $x=\frac{-4}{9},y=\frac{5}{12}$ and $z=\frac{7}{18},$

Find $x+(y+z)$.

Answer 1

Given, $x=\frac{-4}{9},y=\frac{5}{12}andz=\frac{7}{18}$
Here, $x+(y+z)=\frac{-4}{9}+(\frac{5}{12}+\frac{7}{18})=\frac{-4}{9}+\left(\frac{5\times 3+7\times 2}{36}\right)$ [$\because$ LCM of 12 and 18 = 36]
$=\frac{-4}{9}+\left(\frac{15+14}{36}\right)=\frac{-4}{9}+\frac{29}{36}=\frac{-4\times 4+29}{36}=\frac{13}{36}$