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Chapter 7: Problem 61

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

Short Answer

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

Step by step solution

01

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 \(\frac{4x+3}{(x-3)(x+3)}\).
02

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 \(\frac{(x+1)(x+3)}{(x-3)(x+3)}\).
03

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 \(\frac{4x+3-(x^{2}+4x+3)}{(x-3)(x+3)}\).
04

Simplify the result

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

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Most popular questions from this chapter

Describe how to identify the corresponding sides in similar triangles.

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. I used \(\frac{a}{d}=\frac{b}{e}\) to show that corresponding sides of similar triangles are proportional, but I could also use \(\frac{a}{b}=\frac{d}{e}\) or \(\frac{d}{a}=\frac{e}{b}\)

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{2 x-y}{x-y}+\frac{x-2 y}{y-x}$$

add or subtract as indicated. Simplify the result, if possible. $$\frac{6 y^{2}+y}{2 y^{2}-9 y+9}-\frac{2 y+9}{2 y^{2}-9 y+9}-\frac{4 y-3}{2 y^{2}-9 y+9}$$

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{y-3}{y^{2}-25}+\frac{y-3}{25-y^{2}}$$
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