## Elementary Algebra

{$-\frac{8}{5},1$}
We will first cross multiply the fractions together in order to create a quadratic equation: $\frac{5x-2}{3}=\frac{2}{x+1}$ $(5x-2)(x+1)=2(3)$ $5x(x+1)-2(x+1)=6$ $5x^{2}+5x-2x-2=6$ $5x^{2}+3x-2-6=0$ $5x^{2}+3x-8=0$ Now, we will factor this quadratic equation using the rules of factoring trinomials: $5x^{2}+3x-8=0$ $5x^{2}-5x+8x-8=0$ $5x(x-1)+8(x-1)=0$ $(x-1)(5x+8)=0$ $(x-1)=0$ or $(5x+8)=0$ $x=1$ or $x=-\frac{8}{5}$ Therefore, the solution set is {$-\frac{8}{5},1$}.