Problem 51 - Introductory Algebra for College Students

Question

Add or subtract as indicated. Simplify the result, if possible. $$\frac{3}{x^{2}-1}+\frac{4}{(x+1)^{2}}$$

Step-by-Step Solution

Answer

The simplified sum of the two fractions is \( \frac{7x-1 }{(x+1)^{3} * (x-1)} \)

Step 1: Get the Common Denominator

A common denominator between \(x^{2}-1\) and \((x+1)^{2}\) could be found by just multiplying them. So the common denominator is \( x^{2}-1 \) * \( (x+1)^{2} = (x+1)^{2} * (x-1)(x+1) = (x+1)^{3} * (x-1) \).

Step 2: Adjust the Fractions to the Common Denominator

To adjust the fractions to the common denominator, we multiply each term to get the common denominator. For the first term, multiply the numerator and the denominator by \((x+1)\). For the second term, multiply the numerator and the denominator by \((x-1)\). Therefore, \( \frac{3(x+1)}{(x+1)^{3} * (x-1)} + \frac{4(x-1)}{(x+1)^{3} * (x-1)} \).

Step 3: Sum the Fractions

Now that the fractions have the same denominator, we can add the numerators. This gives us \( \frac{3x+3+4x-4}{(x+1)^{3} * (x-1)} \).

Step 4: Simplify the Numerator

We simplify the numerator by combining like terms, which gives us \( \frac{7x-1 }{(x+1)^{3} * (x-1)} \).