Algebra 2 Common Core

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

Chapter 3 - Linear Systems - 3-6 Solving Systems Using Matrices - Practice and Problem-Solving Exercises - Page 181: 56

Answer

The solution is $y = 10$ or $y = -6$.

Work Step by Step

To solve absolute value equations, we get rid of the absolute value sign by setting the expression within the absolute value sign equal to the constant given, but with opposite signs. For this problem, we set up the problem this way: $2y - 4 = 16$ or $2y - 4 = -16$ Solve each equation. First equation: $2y - 4 = 16$ $2y = 4 + 16$ $2y = 20$ Divide both sides of the equation by $2$ to solve for $y$: $y = 10$ Second equation: $2y - 4 = -16$ $2y = 4 - 16$ $2y = -12$ Divide both sides of the equation by $2$ to solve for $y$: $y = -6$ The solutions are $y = 10$ and $y = -6$. Check our answers by plugging them into the original equation to see if it holds true. For $y = 10$: $|2(10) - 4| = 16$ $|20 - 4| = 16$ $|16| = 16$ $16 = 16$ For $y = -6$: $|2(-6) - 4| = 16$ $|-12 - 4| = 16$ $|-16| = 16$ $16 = 16$ The two sides equal one another; therefore; $y = -6$ is also a valid solution.
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