Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter R - Test - Chapter R Test: 22

Answer

$10k-10$

Work Step by Step

We are given the expression $-3(2k-4)+4(3k-5)-2+4k$. We can use the distributive property to simplify the terms in parentheses and then combine like terms. $-3(2k-4)+4(3k-5)-2+4k=(-3\times2k)+(-3\times-4)+(4\times3k)+(4\times-5)-2+4k=(-6k)+(12)+(12k)+(-20)-2+4k=(-6+12+4)k+(12-20-2)=(10)k+(-10)=10k-10$
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