Make x the subject of the given formula $\frac{3}{y} = \sqrt{\frac{y}{2x^2 - 1}}$. Answer x = ...

1. Remove the square root by squaring both sides.

$\frac{3}{y}=\sqrt{\frac{y}{2x^2-1}}$
Squaring both sides,
$\left(\frac{3}{y}\right)^2=\frac{y}{2x^2-1}$
$\frac{9}{y^2}=\frac{y}{2x^2-1}$

2. Clear fractions and rearrange to make $x^2$ the subject.

$\frac{9}{y^2}=\frac{y}{2x^2-1}$
Cross-multiplying,
$9(2x^2-1)=y^3$
$18x^2-9=y^3$
$18x^2=y^3+9$
$x^2=\frac{y^3+9}{18}$

3. Take the square root to make $x$ the subject.

$x=\pm\sqrt{\frac{y^3+9}{18}}$