waecmaths question:
Make p the subject of relation: $q=\frac{3p}{r}+\frac{s}{2}$
Option A:
$p=\frac{2q-rs}{6}$
Option B:
$2qr-sr-3$
Option C:
$p=\frac{2qr-s}{6}$
Option D:
$\frac{2qr-rs}{6}$
waecmaths solution:
\[\begin{align}
& q=\frac{3p}{r}+\frac{s}{2} \\
& \text{Subtract }\frac{s}{2}\text{ from both sides} \\
& q-\frac{s}{2}=\frac{3p}{r} \\
& \frac{2q-s}{2}=\frac{3q}{r} \\
& \text{Multiply both sides by }r \\
& \frac{2q-rs}{2}=3q \\
& \text{Divide both sides by }3 \\
& \frac{2q-rs}{6}=q \\
\end{align}\]
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