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Chapter 7: Problem 69

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

Short Answer

Expert verified
The solution is \(\frac{-x^2 + 7x + 3}{(x-3)(x+2)}\) with \(x\neq 3\) and \(x\neq -2\)

Step by step solution

01

Factoring the denominator of the first fraction

The denominator \(x^{2} - x -6\) of the first fraction can be factored as \( (x-3)(x+2) \). Therefore, the first fraction in simplified form is: \(\frac{9x+3}{(x-3)(x+2)}\).
02

Re-arranging the denominator of the second fraction

The denominator of the second fraction is \(3-x\). The opposite of the factor \(x-3\), which appears in the denominator of the first fraction. Therefore, the second fraction can be rewritten as: \(-\frac{x}{x-3}\). Please note that changing the order within the subtraction, introduces a negative factor.
03

Re-writing the second fraction with common denominator

To make the denominators match, rewrite the second fraction as \(-\frac{x(x+2)}{(x-3)(x+2)}\), introducing the additional factor \((x+2)\). Please note, that this means \(x \neq -2\).
04

Perform addition with common denominator

Now that both fractions share a common denominator, they can be added. Add the numerators together: \(\frac{9x+3}{(x-3)(x+2)} - \frac{x(x+2)}{(x-3)(x+2)} = \frac{9x+3 - x(x+2)}{(x-3)(x+2)}\).
05

Simplified answer

Simplify the numerator and denominator which gives the final solution to be: \(\frac{9x+3 - x^2 - 2x}{(x-3)(x+2)} = \frac{-x^2 + 7x + 3}{(x-3)(x+2)}\). Please consider that both, \(x\neq 3\) and \(x \neq -2\), due to the terms in the denominator.

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