waecmaths question:
Make x the subject of the formula $d=\sqrt{\frac{6}{x}-\frac{y}{2}}$
Option A:
$x=\frac{6}{{{x}^{2}}}+\frac{12}{y}$
Option B:
$\frac{12}{2{{d}^{2}}-y}$
Option C:
$\frac{12}{y}-2{{d}^{2}}$
Option D:
$\frac{12}{2{{d}^{2}}+y}$
waecmaths solution:
$\begin{align} & d=\sqrt{\frac{6}{x}-\frac{y}{2}} \\ & \text{Square both sides of the equation} \\ & {{d}^{2}}=\frac{6}{x}-\frac{y}{2} \\ & \text{Add }\frac{y}{2}\text{ to both sides of the equation} \\ & {{d}^{2}}+\frac{y}{2}=\frac{6}{x} \\ & \frac{2{{d}^{2}}+y}{2}=\frac{6}{x} \\ & \text{Divide both sides by }6 \\ & \frac{2{{d}^{2}}+y}{12}=\frac{1}{x} \\ & \text{Determine the reciprocal of both sides} \\ & x=\frac{12}{2{{d}^{2}}+y} \\\end{align}$
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