what is the solution to the equation $\frac{-3d}{d^{2}-2d - 8}+\frac{3}{d - 4}=\frac{-2}{d + 2}$?\n$d=-4$…

what is the solution to the equation $\frac{-3d}{d^{2}-2d - 8}+\frac{3}{d - 4}=\frac{-2}{d + 2}$?\n$d=-4$ and $d = 2$\n$d=-2$ and $d = 4$\n$d = 1$\n$d = 2$

what is the solution to the equation $\frac{-3d}{d^{2}-2d - 8}+\frac{3}{d - 4}=\frac{-2}{d + 2}$?\n$d=-4$ and $d = 2$\n$d=-2$ and $d = 4$\n$d = 1$\n$d = 2$

Answer

Answer:

d = 1

Explanation:

Step1: Factor the denominator

$d^{2}-2d - 8=(d - 4)(d+2)$

Step2: Find the common denominator

The common denominator of $\frac{-3d}{d^{2}-2d - 8},\frac{3}{d - 4},\frac{-2}{d + 2}$ is $(d - 4)(d + 2)$.

Step3: Rewrite the equation with common - denominator

$\frac{-3d}{(d - 4)(d + 2)}+\frac{3(d + 2)}{(d - 4)(d + 2)}=\frac{-2(d - 4)}{(d - 4)(d + 2)}$

Step4: Simplify the numerators

$-3d+3(d + 2)=-2(d - 4)$

Step5: Expand the expressions

$-3d+3d+6=-2d + 8$

Step6: Combine like - terms

$6=-2d + 8$

Step7: Solve for d

$2d=2$, so $d = 1$