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: 53

Answer

$x=\frac{3}{2}$ $y=-\frac{3}{2}$

Work Step by Step

If we multiply the first equation by $-2$, the second one by $3$, then we add the two equations, we can eliminate the $y$ from the system: $3\times(-2)x+3\times(-2)y=0\times(-2)$ $4\times3x+2\times3y=3\times3$ $-6x+12x=9$ $x=\frac{3}{2}$ $3\times\frac{3}{2}+3\times y=0$ $y=-\frac{3}{2}$ We have to check, if the solution is right. Substitute $x$ with $\frac{3}{2}$ and $y$ with $-\frac{3}{2}$: $3(\frac{3}{2})+3(-\frac{3}{2})=0$ $\checkmark$ $4(\frac{3}{2})+2(-\frac{3}{2})=3$ $\checkmark$
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