Question

Solve the following equations:
$\frac{8\sqrt{x-5}}{3x-7}=\frac{\sqrt{3x-7}}{x-5}$.

Answer 1

$\frac{8\sqrt{x-5}}{3x-7}=\frac{\sqrt{3x-7}}{x-5}$

$8(x-5)\sqrt{x-5}=(3x-7)\sqrt{3x-7}$

Squaring both sides

$64{(x-5)}^3={(3x-7)}^3$

$4^3{(x-5)}^3={(3x-7)}^3$

Taking cube root on both sides, we get

$4(x-5)=3x-7$

$4x-20=3x-7$

$4x-3x=-7+20$

$x=13$