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: 60

Answer

$x=0$ $y=3$

Work Step by Step

If we multiply the second equation by $-3$, then we add the two equations, we can eliminate the $y$ from the system: $\frac{3}{2}x+y=3$ $\frac{2}{3}\times (-3)x+\frac{1}{3}\times(-3)y=1\times (-3)$ $\frac{3}{2}x-2x=0$ $x=0$ $\frac{3}{2}(0)+y=3$ $y=3$ We have to check, if the solution is right. Substitute $x$ with 0 and $y$ with 3: $\frac{3}{2}(0)+3=3$ $\checkmark$ $\frac{2}{3}(0)+\frac{1}{3}(3)=1$ $\checkmark$
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