## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 9 - 9.2 - Two-Variable Linear Systems - 9.2 Exercises - Page 649: 65

#### Answer

a) $(2,-1)$ b) When we solve the system using the elimination method, it becomes easy because $2y$ is common to both equations.

#### Work Step by Step

a) Rewrite the first equation as: $2y=4-3x$ or, $y=2-\dfrac{3x}{2}$ Therefore, the first equation yields: $5x-2(2-\dfrac{3x}{2}) =12 \implies x=2$ Substitute the value of $x$ into the first equation to get the value of $y$. Thus, $y=2-\dfrac{3(2)}{2} \implies y=-1$ So, the solution is $(2,-1)$ b) When we solve the system using the elimination method, it becomes easy because $2y$ is common to both equations.

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