Question

What is the solution to the equation StartFraction negative 3 d Over d squared minus 2 d minus 8 EndFraction + StartFraction 3 Over d minus 4 EndFraction = StartFraction negative 2 Over d + 2 EndFraction? d = –4 and d = 2 d = –2 and d = 4 d = 1 d = 2

Answer 1

Answer:

d = 1

Step-by-step explanation:

-3d/(d²-2d-8) + 3/(d-4) = -2/(d+2)

d²-2d-8 = d²-4d+2d-8

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

= (d+2)(d-4)

Since these the factors in the denominator, they can't be zero.

d cannot be 4 or -2

-3d/[(d+2)(d-4)] + 3/(d-4) = -2/(d+2)

Multiply both sides by: (d+2)(d-4)

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

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

2d = 2

d = 1

Answer 2

The solution to the given equation is 1.

The given parameters:

• $\frac{-3d}{d^2 - 2d + 8} + \frac{3}{d -4} = \frac{-2}{d + 2}$

The solution of the equation is calculated as follows;

$\frac{-3d}{d^2 - 2d + 8} + \frac{3}{d -4} = \frac{-2}{d + 2}\\\\ \frac{3}{d -4} +\frac{2}{d + 2} = \frac{3d}{d^2 - 2d + 8} \\\\\frac{3(d+2) + 2(d-4)}{(d -4)(d+ 2)} = \frac{3d}{d^2 - 2d + 8} \\\\\frac{3(d+2) + 2(d-4)}{d^2 - 2d + 8} =\frac{3d}{d^2 - 2d + 8}$

The denominators are equal, so we can equate the numerators as follows;

$3(d + 2) + 2(d-4) = 3d\\\\3d + 6 + 2d - 8 = 3d\\\\5d - 2 = 3d\\\\5d - 3d = 2\\\\2d = 2\\\\d = 1$

Thus, the solution to the given equation is 1.

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