Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 1 - Expressions, Equations, and Inequalities - Chapter Review - Page 52: 27

Answer

$x = 1$

Work Step by Step

We use the definition that $|x|=a \longrightarrow x=-a \quad \text{or} \quad x=a$ to obtain: $$2x + 8 = 3x + 7\quad \text{or} \quad2x + 8 = - (3x + 7)$$ Solve the first equation first. Subtract $8$ from each side to isolate the constants to one side: $$2x = 3x - 1$$ We now subtract $3x$ from each side to isolate the variable to one side: $$-x = -1$$ Divide each side by $-1$ to isolate the $x$: $$x = 1$$ Let's solv Solve the other equation. Use distributive property on the right side of the equation to obtain: $$2x + 8 = - 3x - 7$$ Subtract $8$ from each side of the equation to isolate the constants to one side: $$2x = - 3x - 15$$ Add $3x$ to each side to isolate the variable to one side of the equation: $$5x = -15$$ Divide both sides of the equation by $5$ to isolate the $x$: $$x = -3$$ Check for extraneous solutions. Plug $x = 1$ into the original equation. $|2(1) + 8| = 3(1) + 7$ Multiply out: $|2 + 8| = 3 + 7$ Add constants on each side of the equation: $|10| = 10$ This statement is true; therefore, 1 is a solution of this equation. Substitute $-3$ for $x$: $|2(-3) + 8| = 3(-3) + 7$ Multiply out: $|-6 + 8| = -9 + 7$ Add constants on each side of the equation: $|2| = -2$ This statement is false; therefore, $-3$ is not a solution of this equation.
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