Problem 61 - Introductory Algebra for College Students

Question

Add or subtract as indicated. Simplify the result, if possible. $$\frac{4 x+3}{x^{2}-9}-\frac{x+1}{x-3}$$

Step-by-Step Solution

Answer

The solution is \(\frac{-x^{2}+3}{(x-3)(x+3)}\)

Step 1: Decompose the denominator of the first fraction

The first fraction's denominator is \(x^{2}-9\), which is the difference of squares and can be factored into \((x-3)(x+3)\). Hence, \(\frac{4 x+3}{x^{2}-9}\) becomes \(\frac{4x+3}{(x-3)(x+3)}\).

Step 2: Find a common denominator

The denominator for the second fraction is \(x-3\) and the first is \((x-3)(x+3)\). We can rewrite the second fraction with a denominator of \((x-3)(x+3)\) by multiplying both the denominator and numerator by \(x+3\). So \(\frac{x+1}{x-3}\) becomes \(\frac{(x+1)(x+3)}{(x-3)(x+3)}\).

Step 3: Subtract the fractions

Now that both fractions have the same denominator \((x-3)(x+3)\), they can be subtracted: \(\frac{4x+3}{(x-3)(x+3)} - \frac{(x+1)(x+3)}{(x-3)(x+3)}\). This simplifies to \(\frac{4x+3-(x^{2}+4x+3)}{(x-3)(x+3)}\).

Step 4: Simplify the result

The term in the numerator simplifies to \(-x^{2}+3\). The final result is \(\frac{-x^{2}+3}{(x-3)(x+3)}\).