Intermediate Algebra (12th Edition)

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

Chapter 3 - Section 3.1 - Systems of Linear Equations in Two Variables - 3.1 Exercises - Page 227: 47

Answer

$x=3$ $y=-1$

Work Step by Step

If we multiply the second equation by $5$, then we add the two equations, we can eliminate the $y$ from the system: $2x-5y=11$ $3\times5x+5y=8\times5$ $2x+15x=51$ $x=3$ $2(3)-5y=11$ $y=-1$ We have to check, if the solution is right. Substitute $x$ with 3 and $y$ with -1: $2(3)-5(-1)=11$ $\checkmark$ $3(3)+(-1)=8$ $\checkmark$
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