Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Section 2.6 - Absolute Value Equations - Exercise Set: 20

Answer

$y = 1$ OR $y = -1/3$

Work Step by Step

$|X| = |Y|$ is true only if $X = Y$ OR $X = -Y$ Given $|9y + 1| = |6y + 4|$ either $9y + 1= 6y + 4$ OR $9y + 1= -(6y + 4)$ IF $9y + 1= 6y + 4$ subtract $6y$ from each side of the equation $3y + 1 = 4$ subtract $1$ from each side of the equation $3y = 3$ divide each side of the equation by $3$ $y = 1$ IF $9y + 1= -(6y + 4)$ distribute the negative on the right side $9y + 1= -6y - 4$ add $6y$ to each side of the equation $15y + 1 = -4$ Subtract $1$ from each side of the equation $15y = -5$ divide each side of the equation by $15$ $y = -5/15$ $y = -1/3$
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