which solution to the equation $\frac{1}{x - 1}=\frac{x - 2}{2x^{2}-2}$ is extraneous?\n$x = 1$ and…

Explanation:

Step1: Factor the denominator

Factor $2x^{2}-2$ as $2(x^{2}-1)=2(x - 1)(x + 1)$. The original equation $\frac{1}{x - 1}=\frac{x - 2}{2x^{2}-2}$ becomes $\frac{1}{x - 1}=\frac{x - 2}{2(x - 1)(x + 1)}$.

Step2: Multiply both sides by the common - denominator

Multiply both sides of the equation by $2(x - 1)(x + 1)$ to get $2(x + 1)=x - 2$.

Step3: Expand and solve for x

Expand the left - hand side: $2x+2=x - 2$. Subtract $x$ from both sides: $2x - x+2=x - x - 2$, which gives $x+2=-2$. Then subtract 2 from both sides: $x=-4$.

Step4: Check for extraneous solutions

The original equation has a denominator of $x - 1$ and $2x^{2}-2 = 2(x - 1)(x + 1)$. A value of $x$ that makes the denominator equal to 0 is an extraneous solution. The denominator $x - 1 = 0$ when $x = 1$. When $x = 1$, the original rational equation is undefined.

Answer:

$x = 1$