Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 3 Linear Systems and Matrices - 3.2 Solve Linear Systems Algebraically - 3.2 Exercises - Skill Practice - Page 165: 49



Work Step by Step

After multiplying the first equation by $6$ we get: $3(x-1)+2(y+2)=24\\3x-3+2y+4=24\\3x+2y=23$ If we multiply the second equation by $-3$ we get: $-3x+6y=-15$. Adding this to the modified version of the first equation we get: $8y=8\\y=1$. Then $3x+2(1)=23\\3x=21\\x=7$
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