What is the answer to 1/((x^2-2x-8))= a/((x-4))+b/((x+2)) ?

The answer to 1/((x^2-2x-8))= a/((x-4))+b/((x+2)) is x=(1-2a+4b)/(a+b);\quad a\ne-b

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1/((x^2-2x-8))= a/((x-4))+b/((x+2))

\frac{1}{( x ^{2} - 2 x - 8 )} = \frac{a}{( x - 4 )} + \frac{b}{( x + 2 )}

x = \frac{1 - 2 a + 4 b}{a + b}; a \neq = - b

Solution steps

\frac{1}{( x ^{2} - 2 x - 8 )} = \frac{a}{( x - 4 )} + \frac{b}{( x + 2 )}

Multiply by LCM

1 = a (x + 2) + b (x - 4)

Switch sides

a (x + 2) + b (x - 4) = 1

Expand

a (x + 2) : a x + 2 a

a x + 2 a + b (x - 4) = 1

Expand

b (x - 4) : b x - 4 b

a x + 2 a + b x - 4 b = 1

Move

2 a

to the right side

a x + b x - 4 b = 1 - 2 a

Move

4 b

to the right side

a x + b x = 1 - 2 a + 4 b

Factor

a x + b x : x (a + b)

x (a + b) = 1 - 2 a + 4 b

Divide both sides by

a + b; a \neq = - b

x = \frac{1 - 2 a + 4 b}{a + b}; a \neq = - b

\frac{1}{x ^{2}} = 0

(\frac{y ^{2}}{2 y}) + 3 y - 5 = 2

\frac{x ^{2} + 1}{x ^{2} + 1} = 0

\frac{1}{5} = \frac{x}{x + 100}

\frac{- 2}{3 ( 2 x - 4 )} = 3 - 2 (x - 8)

What is the answer to 1/((x^2-2x-8))= a/((x-4))+b/((x+2)) ?
The answer to 1/((x^2-2x-8))= a/((x-4))+b/((x+2)) is x=(1-2a+4b)/(a+b);\quad a\ne-b