WAEC 2022 · Paper 2 · Q2

Given that $y = \left(\dfrac{pr}{m} - p^2r\right)^{-\frac32}$ ,

(a)
make $r$ the subject;
Show the answer
$r = \dfrac{m\,y^{-\frac23}}{p - p^2m}$
(b)
find the value of $r$ when $y = -8$ , $m = 1$ and $p = 3$ .

Worked solution (try it first)

(a)

The bracket is raised to the power $-\frac32$ .
Undo it by raising both sides to the power $-\frac23$ : $y^{-\frac23} = \frac{pr}{m} - p^2 r$ .
Multiply every term by $m$ : $m\,y^{-\frac23} = pr - p^2 m r$ .
Take out $r$ : $m\,y^{-\frac23} = r(p - p^2 m)$ .
So $r = \frac{m\,y^{-\frac23}}{p - p^2 m}$ .

(b)

$(-8)^{-\frac23} = \frac{1}{(-8)^{\frac23}}$
$= \frac{1}{\left(\sqrt[3]{-8}\right)^2}$
$= \frac{1}{(-2)^2}$
$= \frac14$ .
With $m = 1$ and $p = 3$ : $r = \frac{1 \times \frac14}{3 - 9 \times 1}$
$= \frac{\frac14}{-6}$
$= -\frac{1}{24}$ .

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