## College Algebra (6th Edition)

$x=-8$
$\dfrac{3}{x+3}=\dfrac{5}{2x+6}+\dfrac{1}{x-2}$ Take out common factor $2$ from the denominator of the first fraction on the right side: $\dfrac{3}{x+3}=\dfrac{5}{2(x+3)}+\dfrac{1}{x-2}$ Multiply the whole equation by $2(x+3)(x-2)$: $2(x+3)(x-2)\Big[\dfrac{3}{x+3}=\dfrac{5}{2(x+3)}+\dfrac{1}{x-2}\Big]$ $(2)(x-2)(3)=(5)(x-2)+(2)(x+3)$ $6x-12=5x-10+2x+6$ Take all terms with $x$ to the left side and all terms without $x$ to the right side and solve for $x$: $6x-5x-2x=-10+6+12$ $-x=8$ $x=-8$