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: 49


$x=2$ $y=-3$

Work Step by Step

If we multiply the first equation by $-3$ second equation by $4$, then we add the two equations, we can eliminate the $y$ from the system: $3x\times(-3)+4\times(-3)y=-6\times(-3)$ $5\times4x+3\times4y=1\times4$ $-9x+20x=22$ $x=2$ $3(2)+4y=-6$ $y=-3$ We have to check, if the solution is right. Substitute $x$ with 2 and $y$ with -3: $3(2)+4(-3)=-6$ $\checkmark$ $5(2)+3(-3)=1$ $\checkmark$
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