Respuesta :
Answer:
d=1
Step-by-step explanation:
[tex]\frac{-3d}{d^2-2d-8} +\frac{3}{d-4} =\frac{-2}{d+2}[/tex]
Lets factor the denominator d^2 -2d-8
d^2 - 2d - 8 = (d-4)(d+2)
[tex]\frac{-3d}{(d-4)(d+2)} +\frac{3}{d-4} =\frac{-2}{d+2}[/tex]
Now make the denominators same
LCD: (d-4)(d+2)
[tex]\frac{-3d}{(d-4)(d+2)} +\frac{3(d+2)}{(d-4)(d+2)} =\frac{-2(d-4)}{(d+2)(d-4)}[/tex]
Denominators are same on both sides
So equate the numerators
-3d +3(d+2) = -2(d-4)
-3d +3d +6 = -2d +8
6 = -2d + 8
subtract 8 on both sides
-2 = -2d
So d=1
