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 86: 23


$\frac{x+1}{3}\div\frac{3x+3}{7}$ $=\frac{x+1}{3}\times\frac{7}{3x+3}$ $=\frac{x+1}{3}\times\frac{7}{3(x+1)}$ $=\frac{1}{3}\times\frac{7}{3}$ $=\frac{7}{9}$

Work Step by Step

To divide rational expressions the first step is to invert the divisor and multiply. This is done by flipping the second fraction, $\frac{3x+3}{7}$ so that it becomes $\frac{7}{3x+3}$ and changing the division to multiplication. The next step is to factorise as many numerators and denominators as possible. In this case $3x+3$ can be factorised to $3(x+1)$ as the common multiple is $3$. The next step is to divide the common numerators and denominators so that they cancel out. $x+1$ is the only common numerator and denominator so they both cancel each other out. Because $x+1$ goes into $x+1$ once the numerator for the first fraction is $1$. The second fraction is left with a denominator of $3$. Lastly the two fractions are simplified to give the final answer.
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