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 3 - Graphing Linear Functions - 3.3 - Function Notation - Exercises - Page 125: 18

Answer

$x=15$

Work Step by Step

$j(x)=-\frac{4}{5}x+7$ Substitute $-5$ for $j(x)$ to get $-5=-\frac{4}{5}x+7$ Subtract $7$ from both sides of the equation. Then, we have $-5-7=-\frac{4}{5}x+7-7$ Or $-12=-\frac{4}{5}x$ Multiply by $-\frac{5}{4}$ on both sides to obtain $-12(-\frac{5}{4})=-\frac{4}{5}x(-\frac{5}{4})$ $\implies 15=x$ When $x=15$, $j(x)=-5$
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