## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 3 - Section 3.1 - Systems of Linear Equations in Two Variables - 3.1 Exercises: 54

#### Answer

$x=\frac{1}{4}$ $y=-\frac{1}{2}$

#### Work Step by Step

If we multiply the second equation by $2$, then we add the two equations, we can eliminate the $y$ from the system: $8x+4y=0$ $4\times2x-2\times2y=2\times2$ $8x+8x=4$ $16x=4$ $x=\frac{1}{4}$ $8\times\frac{1}{4}+4\times y=0$ $y=-\frac{1}{2}$ We have to check, if the solution is right. Substitute $x$ with $\frac{1}{4}$ and $y$ with $-\frac{1}{2}$: $8(\frac{1}{4})+4(-\frac{1}{2})=0$ $\checkmark$ $4(\frac{1}{4})-2(-\frac{1}{2})=2$ $\checkmark$

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