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: 35

Answer

The solutions are $n=5$ and $n=3$.

Work Step by Step

The given equation is $\Rightarrow |4n-15|=|n|$ Write the two related linear equations. $4n-15=n$ ...... (1) $4n-15=-n$ ...... (2) Solve equation (1). $\Rightarrow 4n-15=n$ Add $15-n$ to each side. $\Rightarrow 4n-15+15-n=n+15-n$ Simplify. $\Rightarrow 3n=15$ Divide each side by $3$. $\Rightarrow \frac{3n}{3}=\frac{15}{3}$ Simplify. $\Rightarrow n=5$ Check : $n=5$ $\Rightarrow |4n-15|=|n|$ $\Rightarrow |4(5)-15|=|5|$ $\Rightarrow |20-15|=|5|$ $\Rightarrow |5|=|5|$ True. Solve equation (2). $\Rightarrow 4n-15=-n$ Add $15+n$ to each side. $\Rightarrow 4n-15+15+n=-n+15+n$ Simplify. $\Rightarrow 5n=15$ Divide each side by $5$. $\Rightarrow \frac{5n}{5}=\frac{15}{5}$ Simplify. $\Rightarrow n=3$ Check : $n=3$ $\Rightarrow |4n-15|=|n|$ $\Rightarrow |4(3)-15|=|3|$ $\Rightarrow |12-15|=|3|$ $\Rightarrow |3|=|3|$ True. Hence, the solutions are $n=5$ and $n=3$.
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