Intermediate Algebra (6th Edition)

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

Chapter 2 - Section 2.1 - Linear Equations in One Variable - Exercise Set - Page 56: 66

Answer

$n=-\frac{2}{3}$

Work Step by Step

$3[8-4(n-2)]+5n=-20+2[5(1-n)-6n]$ $3[8-4n+8]+5n=-20+2[5-5n-6n]$ $-12n+5n=-20+2[5-11n]$ $-7n=-20+10-22n$ $15n=-10$ $n=-\frac{2}{3}$
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