## Calculus with Applications (10th Edition)

$\dfrac{4}{2x^{2}+3x-9}+\dfrac{2}{2x^{2}-x-3}=\dfrac{3}{x^{2}+4x+3}$ Factor all three rational expressions completely: $\dfrac{4}{(x+3)(2x-3)}+\dfrac{2}{(x+1)(2x-3)}=\dfrac{3}{(x+3)(x+1)}$ Multiply the whole equation by $(x+3)(2x-3)(x+1)$: $(x+3)(2x-3)(x+1)\Big[\dfrac{4}{(x+3)(2x-3)}+\dfrac{2}{(x+1)(2x-3)}=\dfrac{3}{(x+3)(x+1)}\Big]$ $4(x+1)+2(x+3)=3(2x-3)$ $4x+4+2x+6=6x-9$ Take all terms to the left side and simplify: $4x+4+2x+6-6x+9=0$ $19\ne0$ False Since simplifying resulted in a false statement, this equation has no solution.