Question

question
solve for all values of x.
1 - \frac{2}{x - 1}=\frac{5}{x^{2}-1}
answer
attempt 1 out of 2
additional solution no solution
x =

Explanation:

Step1: Factor the denominator

Note that $x^{2}-1=(x + 1)(x - 1)$. The given equation $1-\frac{2}{x - 1}=\frac{5}{x^{2}-1}$ can be rewritten with a common - denominator.

Step2: Multiply through by the common denominator

Multiply each term in the equation by $(x + 1)(x - 1)$:
$(x + 1)(x - 1)\times1-(x + 1)(x - 1)\times\frac{2}{x - 1}=(x + 1)(x - 1)\times\frac{5}{(x + 1)(x - 1)}$
$(x + 1)(x - 1)-2(x + 1)=5$
Expand the expressions: $x^{2}-1-2x - 2 = 5$

Step3: Simplify the equation

Combine like - terms: $x^{2}-2x-3 = 5$
Move all terms to one side to get a quadratic equation: $x^{2}-2x - 8=0$

Step4: Solve the quadratic equation

Factor the quadratic equation: $x^{2}-2x - 8=(x - 4)(x+2)=0$
Set each factor equal to zero:
If $x - 4 = 0$, then $x = 4$; if $x + 2=0$, then $x=-2$

Step5: Check for extraneous solutions

We need to check if these solutions make the original denominators equal to zero.
For the original equation, the denominators are $x - 1$ and $x^{2}-1=(x + 1)(x - 1)$.
When $x = 4$, $x-1=3
eq0$ and $x^{2}-1 = 15
eq0$.
When $x=-2$, $x - 1=-3
eq0$ and $x^{2}-1 = 3
eq0$.

Answer:

$x = 4,x=-2$