## Algebra 1: Common Core (15th Edition)

$\frac{d}{d+2}-\frac{2}{2-d}=\frac{d+6}{d^2-4}$ $\frac{d}{d+2}+\frac{2}{d-2}=\frac{d+6}{(d-2)(d+2)}$ $(d+2)(d-2)(\frac{d}{d+2}+\frac{2}{d-2})=(d+2)(d-2)\frac{d+6}{(d-2)(d+2)}$ $d(d-2)+2(d+2)=d+6$ $d^2-2d+2d+4=d+6$ $d^2-d-2=0$ $d^2-2d+d-2=0$ $d(d-2)+(d-2)=0$ $(d+1)(d-2)=0$ $d+1=0$ or $d-2=0$ $d=-1$ or $d=2$ Check: $-1/(-1+2)-2/[2-(-1)]=-1-2/3=-5/3; (-1+6)/[(-1)^2-4]=5/-3=-5/3$ $2/(2+2)-2(2-2)=2/4-2/0$ (undefined) The equation has one solution, -1.