Question

Take
(iii) $-\frac{3}{2}p+q-r$ from $\frac{1}{2}p-\frac{1}{3}q-\frac{3}{2}r$

Answer 1

Step: Find the value
To subtract: $-\frac{3}{2}p+q-r$ from $\frac{1}{2}p-\frac{1}{3}q-\frac{3}{2}r$

$\therefore (\frac{1}{2}p-\frac{1}{3}q-\frac{3}{2}r)-(-\frac{3}{2}p+q-r)$

$=\frac{1}{2}p-\frac{1}{3}q-\frac{3}{2}r+\frac{3}{2}p-q+r$

$=(\frac{1}{2}p+\frac{3}{2}p)+(-\frac{1}{3}q-q)+(-\frac{3}{2}r+r)$

$=\frac{p+3p}{2}+\frac{(-q-3q)}{3}+\frac{(-3r+2r)}{2}$

$=\frac{4p}{2}+\frac{(-4q)}{3}+\frac{(-r)}{2}$

$=2p-\frac{4}{3}q-\frac{1}{2}r$

Hence, the required value is $2p-\frac{4}{3}q-\frac{1}{2}r.$