## Precalculus (6th Edition) Blitzer

Consider the provided inequality, $\frac{x-2}{x+3}<2$. Multiply both sides by ${{\left( x+3 \right)}^{2}}$ such that, $x\ne -3$ ${{\left( x+3 \right)}^{2}}\frac{x-2}{x+3}<2{{\left( x+3 \right)}^{2}}$ Solve further to get, \begin{align} & \left( x+3 \right)\left( x-2 \right)<2{{\left( x+3 \right)}^{2}} \\ & \left( x-2 \right)<2\left( x+3 \right) \\ & x-2<2x+6 \\ & x>8 \end{align} Hence, the provided statement is true.