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 - Exercises - Page 33: 39

Answer

The solutions are $p=10$ and $p=\frac{2}{3}$.

Work Step by Step

The given equation is $\Rightarrow 4|p-3|=|2p+8|$ Write the two related linear equations. $4(p-3)=2p+8$ ...... (1) $4(p-3)=-(2p+8)$ ...... (2) Solve equation (1). $\Rightarrow 4(p-3)=2p+8$ Use distributive property. $\Rightarrow 4p-12=2p+8$ Add $12-2p$ to each side. $\Rightarrow 4p-12+12-2p=2p+8+12-2p$ Simplify. $\Rightarrow 2p=20$ Divide each side by $2$. $\Rightarrow \frac{2p}{2}=\frac{20}{2}$ Simplify. $\Rightarrow p=10$ Check : $p=10$ $\Rightarrow 4|p-3|=|2p+8|$ $\Rightarrow 4|10-3|=|2(10)+8|$ $\Rightarrow 4|7|=|20+8|$ $\Rightarrow 4(7)=|28|$ $\Rightarrow 28=28$ True. Solve equation (2). $\Rightarrow 4(p-3)=-(2p+8)$ Use distributive property. $\Rightarrow 4p-12=-2p-8$ Add $12+2p$ to each side. $\Rightarrow 4p-12+12+2p=-2p-8+12+2p$ Simplify. $\Rightarrow 6p=4$ Divide each side by $6$. $\Rightarrow \frac{6p}{6}=\frac{4}{6}$ Simplify. $\Rightarrow p=\frac{2}{3}$ Check : $p=\frac{2}{3}$ $\Rightarrow 4|p-3|=|2p+8|$ $\Rightarrow 4|\frac{2}{3}-3|=|2(\frac{2}{3})+8|$ $\Rightarrow 4|\frac{2-9}{3}|=|\frac{4}{3}+8|$ $\Rightarrow 4|\frac{-7}{3}|=|\frac{4+24}{3}|$ $\Rightarrow 4(\frac{7}{3})=|\frac{28}{3}|$ $\Rightarrow \frac{28}{3}=\frac{28}{3}$ True. Hence, the solutions are $p=10$ and $p=\frac{2}{3}$.
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