Answer:
[tex] \boxed{x = \frac{p + q {r}^{2} }{a - b {r}^{2} } } [/tex]
Step-by-step explanation:
[tex]Solve \: for \: x: \\ = > r = \sqrt{ \frac{ax - p}{q + bx} } \\ \\
r = \sqrt{ \frac{ax - p}{q + bx} } \: is \: equivalent \: to \: \sqrt{ \frac{ax - p}{q + bx} } = r :\\ = > \sqrt{ \frac{ax - p}{q + bx} } = r \\ \\ Raise \: both \: sides \: to \: the \: power \: of \: two: \\ = > \frac{ax - p}{q + bx} = {r}^{2} \\ \\ Multiply \: both \: sides \: by \: (q + b x): \\ = > ax - p = {r}^{2} (q + bx) \\ \\ Expand \: out \: terms \: of \: the \: right \: hand \: side: \\ = > ax - p = q {r}^{2} + b {r}^{2}x \\ \\ Subtract \: b {r}^{2}x - p \: from \: both \: sides: \\ = > x(a - b {r}^{2} ) = p + q {r}^{2} \\ \\ Divide \: both \: sides \: by \: a - b {r}^{2} : \\ = > x = \frac{p + q {r}^{2} }{a - b {r}^{2} } [/tex]
