Question

solve: \\(\frac{2}{x - 1} = \frac{4}{x^2 - 1}\\)\\(x = \square\\)enter your answer as a reduced fraction or an integer. enter dne for does not exist, \\(\infty\\) for infinityquestion help: \\(\text{message instructor}\\)question 6

Explanation:

Step1: Factor the denominator

Notice that \(x^2 - 1\) is a difference of squares, so \(x^2 - 1=(x - 1)(x + 1)\). The equation becomes \(\frac{2}{x - 1}=\frac{4}{(x - 1)(x + 1)}\).

Step2: Multiply both sides by \((x - 1)(x + 1)\) (note \(x

eq1\) and \(x
eq - 1\) to avoid division by zero)
\(2(x + 1)=4\).

Step3: Solve the linear equation

Expand the left side: \(2x+2 = 4\). Subtract 2 from both sides: \(2x=4 - 2=2\). Divide both sides by 2: \(x = 1\). But we had a restriction \(x
eq1\) from the original equation (since when \(x = 1\), the denominator \(x - 1=0\) and \(x^2-1 = 0\)). So the solution does not exist.

Answer:

DNE