Make the variable in square brackets the subject: $x=\sqrt{\dfrac{3s-a}{s-b}}$ [$s$].[3]

Assume $x\ge0$, $x^2\ne3$ and $a\ne3b$.

Question

Make the variable in square brackets the subject: $x=\sqrt{\dfrac{3s-a}{s-b}}$ [$s$]. Assume $x\ge0$, $x^2\ne3$ and $a\ne3b$.

Accepted Answer

$s=\dfrac{x^2b-a}{x^2-3}$.

Changing the subject of formulae