Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 27



Work Step by Step

$\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{1}{3x+3}$ First, we can take out common factor $3$ in the denominator of the fraction $\dfrac{1}{3x+3}$ $\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{1}{3(x+1)}$ We multiply the whole equation by $6(x+1)$, which is the least common multiple of the denominators present in the equation. We do this to avoid working with fractions. $6(x+1)[\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{1}{3(x+1)}]$ Evaluate the products and simplify: $\dfrac{18(x+1)}{x+1}-\dfrac{6(x+1)}{2}=\dfrac{6(x+1)}{3(x+1)}$ $18-3(x+1)=2$ $18-3x-3=2$ Solve for $x$: $-3x=2-18+3$ $-3x=-13$ $x=\dfrac{-13}{-3}=\dfrac{13}{3}$
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