Question

Question 94 (h)

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

Find $(x-y)+z$.

Answer 1

Given, $x=\frac{-4}{9},y=\frac{5}{12}$ and $z=\frac{7}{18}$
We have,
$(x-y)+z=(\frac{-4}{9}-\frac{5}{12})+\frac{7}{18}=\frac{-4\times 4-5\times 3}{36}+\frac{7}{18}$ [$\because$ LCM of 9 and 12 = 36]
$=\frac{-16-15}{36}+\frac{7}{18}=(\frac{-31}{36}+\frac{7}{18})$
$=\frac{-31+7\times 2}{36}=\frac{-31+14}{36}$ [$\because$ LCM of 18 and 36 = 36]
$=\frac{-17}{36}$