College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 88: 94

Answer

By finding the least common denominator (LCD) of each rational expression or finding the LCD of all the rational expressions in the numerator and denominator, the simplification $\frac{3x+2}{2x+1}$ can be obtained.

Work Step by Step

1) Finding the LCD of the numerator and denominator of the entire expression separately: Start by finding the LCD of the rational expression in the numerator, $\frac{3}{x}$+$\frac{2}{x^{2}}$, of the whole expression. There is currently an $x$ and $x^{2}$ in the denominators of the rational expression. Therefore, the LCD would be $x^{2}$, as $x$ $\times$ $x$ would give you $x^{2}$. Multiply the numerator and denominator of the first term, $\frac{3}{x}$, by $x$ to get $\frac{3x}{x^{2}}$. The new expression comes to $\frac{3x}{x^{2}}$+$\frac{2}{x^{2}}$, which simplifies to $\frac{3x+2}{x^{2}}$. Repeat this process for the denominator, $\frac{1}{x^{2}}$+$\frac{2}{x}$, of the entire expression. By multiplying both the numerator and denominator of the second term by $x$, the new rational expression becomes $\frac{1}{x^{2}}$+$\frac{2x}{x^{2}}$ , which simplifies to $\frac{1+2x}{x^{2}}$. The entire rational expression now looks like $\frac{\frac{3x+2}{x^{2}}}{\frac{1+2x}{x^{2}}}$. Finish simplifying by multiplying the numerator by the reciprocal of the denominator, so that the $x^{2}$ cancels out and the final answer becomes $\frac{3x+2}{2x+1}$. 2) Finding the LCD of the whole expression: The only denominators in both the numerator and denominator of the entire expression are $x$ and $x^{2}$. Therefore, the LCD would be $x^{2}$, as $x$ $\times$ $x$ would give you $x^{2}$. Multiply the entire expression by $x^{2}$. Distribute: $x^{2}$$(\frac{\frac{3}{x}+\frac{2}{x^{2}}}{\frac{1}{x^{2}}+\frac{2}{x}}$)=$\frac{3x+2}{2x+1}$.
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