Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.4 Adding and Subtracting Rational Expressions - 7.4 Exercises - Page 590: 58

Answer

$\frac{3}{x - 5}$

Work Step by Step

To add or subtract rational expressions, we need to make sure that the expressions have the same denominator, and they do have the same denominator. We need to make sure that each denominator is factored completely. We see that the denominators are quadratic expressions, which are given by the formula: $ax^2 + bx + c$, where $a$, $b$, and $c$ are all real numbers. To factor the expression in the denominator of the first fraction, we want to find which factors when multiplied will give us the product of the $a$ and $c$ terms, which is $-55$, but when added together will give us the $b$ term, which is $6$. This means that one factor is positive and the other negative, with the positive factor having the greater absolute value. Let's look at possible factors: $11$ and $-5$ $55$ and $-1$ It looks like the first combination will work. Let's split the middle term: $x^2 + 11x - 5x - 55$ Group the first two terms and the last two terms: $(x^2 + 11x) - (5x + 55)$ Factor out what is common in both groups: $x(x + 11) - 5(x + 11)$ Group the factors: $(x - 5)(x + 11)$ The exercise can now be rewritten as: $\frac{5x + 43}{(x - 5)(x + 11)} - \frac{2x + 10}{(x - 5)(x + 11)}$ Rewrite the two fractions as one with the same denominator: $\frac{(5x + 43) - (2x + 10)}{(x - 5)(x + 11)}$ Combine like terms in the numerator: $\frac{3x + 33}{(x - 5)(x + 11)}$ Factor out what is common in the numerator: $\frac{3(x + 11)}{(x - 5)(x + 11)}$ Cancel the $x + 11$ factor in the numerator and denominator: $\frac{3}{x - 5}$
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