How do you solve $\frac{7}{x + 1} + \frac{x}{x^{2} - 1} = \frac{1}{x - 1}$ $\frac { 7} { x + 1} + \frac { x } { x ^ { 2} - 1} = \frac { 1} { x - 1}$ ?

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Answer 1 · 2017-08-28T05:16:14

$x = \frac{8}{7}$ $x=8/7$

First, we can eliminate the denominators. If you notice the denominators, the one denominator (the middle) is the product of the other two.

$(x + 1) (x - 1) = x^{2} - 1$ $(x+1)(x-1)=x^2-1$

So, we can multiply through by $(x + 1) (x - 1)$ $(x+1)(x-1)$ .

$\frac{7}{x + 1} + \frac{x}{x^{2} - 1} = \frac{1}{x - 1} \to$ $7/(x+1)+x/(x^2-1)=1/(x-1) ->$

$(x + 1) (x - 1) (\frac{7}{x + 1} + \frac{x}{x^{2} - 1} = \frac{1}{x - 1}) \to$ $(x+1)(x-1)(7/(x+1)+x/(x^2-1)=1/(x-1)) ->$

$7 (x - 1) + x = x + 1$ $7(x-1)+x=x+1$

Then, we can solve for x.

$7 (x - 1) + x = x + 1 \to$ $7(x-1)+x=x+1 ->$

$8 x - 7 = x + 1 \to$ $8x-7=x+1 ->$

$7 x = 8$ $7x=8$

$x = \frac{8}{7}$ $x=8/7$

If you want, you can plug this answer in to check if it is correct.

How do you solve 7x+1+xx2−1=1x−1\frac { 7} { x + 1} + \frac { x } { x ^ { 2} - 1} = \frac { 1} { x - 1}?