Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.4 - Solving Linear Equations - Exercise Set - Page 50: 36



Work Step by Step

For easier work, it is better to get rid of the fractions. This can be achieved by multiplying the LCD of $24$ on both sides of the equation to obtain: $24(5+\frac{x-2}{3}) = 24(\frac{x+3}{8}) \\24(5) + \frac{24(x-2)}{3} = \frac{24(x+3)}{8} \\120+8(x-2) = 3(x+3) \\120+8x-16 = 3x+9 \\8x+104=3x+9$ Subtract $3x$ and $104$ on both sides to obtain: $8x-3x=9-104 \\5x=-95 \\\frac{5x}{5} = \frac{-95}{5} \\x = -19$ Check: $\begin{array}{ccc} &5+\frac{-19-2}{3} &= &\frac{-19+3}{8} \\&5+\frac{-21}{3} &= &\frac{-16}{8} \\&5+(-7) &= &-2 \\&-2 &= &-2\end{array}$
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