Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
rational equations
which solution to the equation \\(\frac{1}{x - 1} = \frac{x - 2}{2x^2 - 2}\\) is extraneous?
options:
\\(x = -4\\)
\\(x = 1\\)
\\(x = 1\\) and \\(x = -4\\)
neither \\(x = 1\\) or \\(x = -4\\)

Explanation:

Step1: Simplify the equation

First, factor the denominators. The right - hand side denominator \(2x^{2}-2 = 2(x^{2}-1)=2(x - 1)(x + 1)\) and the left - hand side denominator is \(x - 1\).
Multiply both sides of the equation \(\frac{1}{x - 1}=\frac{x - 2}{2x^{2}-2}\) by \(2(x - 1)(x + 1)\) (the least common denominator) to get rid of the fractions:
\(2(x + 1)=x - 2\)

Step2: Solve the linear equation

Expand the left - hand side: \(2x+2=x - 2\)
Subtract \(x\) from both sides: \(2x - x+2=x - x - 2\), which gives \(x + 2=-2\)
Subtract 2 from both sides: \(x=-2 - 2=-4\)

Step3: Check for extraneous solutions

Now, we need to check the domain of the original equation. The denominators \(x - 1
eq0\) (so \(x
eq1\)) and \(2x^{2}-2
eq0\), which means \(x^{2}-1
eq0\), so \(x
eq\pm1\).
We found \(x = - 4\), and \(-4\) is not in the excluded values (\(x
eq\pm1\)), so \(x=-4\) is a valid solution.
Now, let's check \(x = 1\). If we substitute \(x = 1\) into the original equation, the denominator of the left - hand side \(\frac{1}{x - 1}\) becomes \(\frac{1}{1 - 1}=\frac{1}{0}\), which is undefined. So \(x = 1\) is an extraneous solution.

Answer:

\(x = 1\)