Sure! Let’s go through the detailed steps to solve the equation:
[tex]\[ \frac{1}{2x + 2} = \frac{1}{x^2 - 1} \][/tex]
Step 1: Recognize and simplify any obvious expressions.
First, let's simplify the denominators on both sides of the equation:
[tex]\[ 2x + 2 \text{ and } x^2 - 1. \][/tex]
Notice that \(x^2 - 1\) is a difference of squares and can be factored as:
[tex]\[ x^2 - 1 = (x + 1)(x - 1) \][/tex]
Step 2: Set the equation with the simplified forms:
[tex]\[ \frac{1}{2(x + 1)} = \frac{1}{(x + 1)(x - 1)} \][/tex]
Step 3: To eliminate the fractions, clear the denominators by multiplying both sides by \( 2(x + 1)(x - 1) \):
[tex]\[ 2(x + 1)(x - 1) \left(\frac{1}{2(x + 1)}\right) = 2(x + 1)(x - 1) \left(\frac{1}{(x + 1)(x - 1)}\right) \][/tex]
This simplifies to:
[tex]\[ (x - 1) = 2 \][/tex]
Step 4: Solve for \( x \) by isolating the term:
[tex]\[ x - 1 = 2 \][/tex]
Add 1 to both sides of the equation:
[tex]\[ x = 3 \][/tex]
Step 5: Check the solution to ensure it's valid.
Substitute \( x = 3 \) back into the original equation to check:
Original equation:
[tex]\[ \frac{1}{2x + 2} = \frac{1}{x^2 - 1} \][/tex]
Substitute \( x = 3 \):
Left-hand side:
[tex]\[ \frac{1}{2(3) + 2} = \frac{1}{6 + 2} = \frac{1}{8} \][/tex]
Right-hand side:
[tex]\[ \frac{1}{3^2 - 1} = \frac{1}{9 - 1} = \frac{1}{8} \][/tex]
Since both sides are equal, \( x = 3 \) is indeed a solution.
So, the solution to the equation \(\frac{1}{2x + 2} = \frac{1}{x^2 - 1}\) is:
[tex]\[ x = 3 \][/tex]
