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: 37

Answer

$x=\dfrac{46}{5}$

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 $21$ on both sides of the equation to obtain: $21(\frac{x+1}{3}) = 21(5-\frac{x+2}{7}) \\\frac{21(x+1)}{3} = 21(5) - \frac{21(x+2)}{7} \\7(x+1) = 105-3(x+2) \\7x+7 = 105-3x-6 \\7x+7=-3x+99$ Add $3x$ and subtract $7$ on both sides to obtain: $7x+3x=99-7 \\10x=92 \\\frac{10x}{10} = \frac{92}{10} \\x = \frac{46}{5}$ Check: $\begin{array}{ccc} &\dfrac{\frac{46}{5}+1}{3} &= &5-\dfrac{\frac{46}{5}+2}{7} \\&\dfrac{\frac{46}{5} + \frac{5}{5}}{3} &= &5-\dfrac{\frac{46}{5} + \frac{10}{5}}{7} \\&\dfrac{\frac{51}{5}}{3} &= &5-\dfrac{\frac{56}{5}}{7} \\&\frac{51}{5(3)}&= &5-\frac{56}{5(7)} \\&\frac{17}{5} &= &5-\frac{8}{5} \\&\frac{17}{5} &= &\frac{25}{5} -\frac{8}{5} \\&\frac{17}{5} &= &\frac{17}{5}\end{array}$
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