What is the answer to 6(1-x^2)^{-1}+3(1-x)^{-1}=5(x+1)^{-1} ?

The answer to 6(1-x^2)^{-1}+3(1-x)^{-1}=5(x+1)^{-1} is x=-1/2

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6(1-x^2)^{-1}+3(1-x)^{-1}=5(x+1)^{-1}

6 (1 - x^{2})^{- 1} + 3 (1 - x)^{- 1} = 5 (x + 1)^{- 1}

x = - \frac{1}{2}

Decimal Notation

x = - 0.5

Solution steps

6 (1 - x^{2})^{- 1} + 3 (1 - x)^{- 1} = 5 (x + 1)^{- 1}

Refine

\frac{6}{1 - x ^{2}} + \frac{3}{1 - x} = \frac{5}{x + 1}

Multiply by LCM

6 (x - 1) + 3 (x + 1) (x - 1) = 5 (x - 1) (- x + 1)

Solve

6 (x - 1) + 3 (x + 1) (x - 1) = 5 (x - 1) (- x + 1) : x = 1, x = - \frac{1}{2}

x = 1, x = - \frac{1}{2}

Verify Solutions

Find undefined (singularity) points:

x = - 1, x = 1

Since the equation is undefined for:

1

x = - \frac{1}{2}

\frac{3}{5} + \frac{7}{6 x} = \frac{1}{8 x}

\frac{1}{x} = \frac{2}{5 x} - \frac{6}{5}

x = \frac{25}{x}

\frac{a - x}{a + x}

a, \frac{4 d}{a} - 9 = g

What is the answer to 6(1-x^2)^{-1}+3(1-x)^{-1}=5(x+1)^{-1} ?
The answer to 6(1-x^2)^{-1}+3(1-x)^{-1}=5(x+1)^{-1} is x=-1/2
Solve for x: 6(1-x^2)^{-1}+3(1-x)^{-1}=5(x+1)^{-1}
The solution is x=-1/2