Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set: 11

Answer

$\dfrac{33}{2} = x$

Work Step by Step

Solving the given equation will be easier if there are no fractions involved. To eliminate the fractions, multiply the LCD of the denominators (which is 24) on both sides of the equation to obtain: $24 \left(\dfrac{x+3}{6}\right) = 24 \left(\dfrac{3}{8} + \dfrac{x-5}{4}\right)$ Distribute 24 on the right side to obtain: $24 \left(\dfrac{x+3}{6}\right) = 24 \left(\dfrac{3}{8}\right) + 24 \left(\dfrac{x-5}{4}\right)$ Cancel the common factors to obtain: $\require{cancel}\begin{array}{ccc} &\cancel{24}4 \left(\dfrac{x+3}{\cancel{6}}\right) &= &\cancel{24}3 \left(\dfrac{3}{\cancel{8}}\right) + \cancel{24}6 \left(\dfrac{x-5}{\cancel{4}}\right) \\&4(x+3) &= &3(3) + 6(x-5)\end{array}$ Use the distributive property to obtain: $4(x) + 4(3) = 9 + 6(x) - 6(5) \\4x+12 = 9 + 6x-30 \\4x+12 = 6x-21$ Subtract $4x$ on both sides of the equation to obtain: $12 = 2x - 21$ Add 21 on both sides of the equation to obtain: $33= 2x$ Divide 2 on both sides of the equation to obtain: $\dfrac{33}{2} = x$
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