Problem 84 - Introductory Algebra for College Students

Question

Perform the indicated operation or operations. Simplify the result, if possible. $$\frac{x+8}{x^{2}-9}-\frac{x+2}{x+3}+\frac{x-2}{x-3}$$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{8}{x^{2}-9}\).

Step 1: Factorization

First, the denominator in the first fraction \(x^{2}-9\) can be factored since it is a difference of squares. The factored form should be \((x-3)(x+3)\). So the first fraction can be rewritten as \(\frac{x+8}{(x-3)(x+3)}\). The entire expression now looks like this: \(\frac{x+8}{(x-3)(x+3)}-\frac{x+2}{x+3}+\frac{x-2}{x-3}\)

Step 2: Finding a common denominator

Next, find a common denominator for all fractions. It can be observed that the common denominator should be \((x-3)(x+3)\) because each denominator can evenly divide into it. Multiply the second and the third fractions by the missing factor to get a common denominator. The expression will look like this: \(\frac{x+8}{(x-3)(x+3)}-\frac{(x+2)(x-3)}{(x-3)(x+3)}+\frac{(x+2)(x+3)}{(x-3)(x+3)}\)

Step 3: Distribute and simplify numerators

Now distribute any terms in the numerators as needed. The expression should look like this: \(\frac{x+8-(x^{2}-x-6)+(x^{2}+x-6)}{(x-3)(x+3)}\)

Step 4: Simplify the expression

Put the expression together and simplify the numerator. Write the numerator as one polynomial and combine like terms. The simplified expression will be \(\frac{8}{(x-3)(x+3)}\)