Question

$\frac{a}{x-8} +\frac{b}{x+4} =\frac{2x-64}{(x-8)(x+4)}$
solve for a and b

Answer 1

Answer:

a = -4

b = 6

Step-by-step explanation:

See attached

Answer 2

Answer:

a=-4 and b=6

Step-by-step explanation:

$\frac{a}{x-8} +\frac{b}{x+4} =\frac{2x-64}{(x-8)(x+4)}$

First, add the fractions by finding the common denominator.

In this case, (x-8)(x+4).

$\frac{a(x+4) + b(x-8)}{(x-8)(x+4)} =\frac{2x-64}{(x-8)(x+4)}$

Therefore, the numerators are equal:

$a(x+4) + b(x-8) =2x-64$

Simplify:

$ax+4a + bx-8b =2x-64\\(a+b)x+4a-8b=2x-64$

Now match the coefficients.

$a+b=2, 4a-8b=-64$

Solve the system of equations.  I'll use substitution, but you can also use elimination if you prefer.

$4a-8b=-64\\a-2b=-16\\(2-b)-2b=-16\\2-3b=-16\\-3b=-18\\b=6\\a=-4$

Therefore, a=-4 and b=6.