Question 81887
cross multiply, then you get
{{{x(x+7)=5(x+3)}}} multiply through
{{{x^2+7x=5x+15}}} get all the terms on one side
{{{x^2+7x-5x-15=0}}} simplify
{{{x^2+2x-15=0}}} this factors into
{{{(x+5)(x-3)=0}}} so 
{{{x+5=0}}} and {{{x-3=0}}} so you have two solutions
{{{x=-5}}} or {{{x=3}}}