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 - 1.3 - Solving Equations with Variables on Both Sides - Monitoring Progress - Page 20: 2

Answer

$h=\frac{3}{8}$

Work Step by Step

$$\begin{align*} \frac{1}{2}(6h)-\frac{1}{2}(4)&=-5h+1 &\text{Apply the Distributive Property.}\\ 3h-2&=-5h+1 &\text{Simplify.}\\ 3h-2+2&=-5h+1+2 &\text{Add $2$ to both sides.}\\ 3h&=-5h+3 &\text{Simplify.}\\ 3h+5h&=-5h+3+5h &\text{Add $5h$ to both sides.}\\ 8h&=3 &\text{Simplify.}\\ \frac{8h}{8}&=\frac{3}{8} &\text{Divide $8$ to both sides.}\\ h&=\frac{3}{8} &\text{Simplify.} \end{align*}$$ Check if the answer is correct by substituting $\frac{3}{8}$ to the variable in the given equation. $$\begin{align*} \frac{1}{2}(6h-4)&=-5h+1\\ \frac{1}{2}\left(6\cdot \frac{3}{8}-4\right)&=-5\cdot \frac{3}{8}+1\\ \frac{1}{2}\left(\frac{-7}{4}\right)&\stackrel{?}{=}\frac{-15}{8}+1\\ \frac{-7}{8}&\stackrel{\checkmark}{=}\frac{-7}{8} \end{align*}$$
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