Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 1 - Solving Linear Equations - 1.4 - Solving Absolute Value Equations - Monitoring Progress - Page 30: 9

Answer

$x=17$ and $x=\frac{7}{5}$

Work Step by Step

The given equation is $\Rightarrow 3|x-4|=|2x+5|$ Write related linear equations. $\Rightarrow 3(x-4)=2x+5$ or $3(x-4)=−(2x+5)$ Solve the first equation. $\Rightarrow 3(x-4)=2x+5$ Use distributive property. $\Rightarrow 3x-12=2x+5$ Add $-2x+12$ to each side. $\Rightarrow 3x-12-2x+12=2x+5-2x+12$ Simplify. $\Rightarrow x=17$ Solve the second equation. $\Rightarrow 3(x-4)=-(2x+5)$ Use distributive property. $\Rightarrow 3x-12=-2x-5$ Add $2x+12$ to each side. $\Rightarrow 3x-12+2x+12=-2x-5+2x+12$ Simplify. $\Rightarrow 5x=7$ Divide each side by $5$. $\Rightarrow \frac{5x}{5}=\frac{7}{5}$ Simplify. $\Rightarrow x=\frac{7}{5}$ Check $x=17$ $\Rightarrow 3|17-4|=|2(17)+5|$ $\Rightarrow 3|13|=|34+5|$ $\Rightarrow 3(13)=|39|$ $\Rightarrow 39=39$ True. Check $x=-3$ $\Rightarrow 3|\frac{7}{5}-4|=|2(\frac{7}{5})+5|$ $\Rightarrow 3|\frac{7-20}{5}|=|\frac{14}{5}+5|$ $\Rightarrow 3|-\frac{13}{5}|=|\frac{14+25}{5}|$ $\Rightarrow 3(\frac{13}{5})=|\frac{39}{5}|$ $\Rightarrow \frac{39}{5}=\frac{39}{5}$ True. Hence, the solutions are $x=17$ and $x=\frac{7}{5}$.
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