## College Algebra (10th Edition)

Published by Pearson

# Chapter 8 - Section 8.1 - Systems of Linear Equations: Substitution and Elimination - 8.1 Assess Your Understanding - Page 555: 29

#### Answer

$(\displaystyle \frac{3}{2},3)$

#### Work Step by Step

Eliminate y, $\left\{\begin{array}{lllll} 2x & -y & =0 & /\times 2 & \\ 4x & +2y & =12 & & ...add \end{array}\right.$ $8x=12$ $x=\displaystyle \frac{12}{8}=\frac{3}{2}$ Back - substitute: $2x-y=0$ $2(\displaystyle \frac{3}{2})-y=0$ $3-y=0$ $y=3$ Solution: $(\displaystyle \frac{3}{2},3)$ Graphing, use x- and y-intercepts. If the origin is the intercept, select another x and calculate y $\left[\begin{array}{lll} & & \\ & (y=0) & (x=1)\\ 2x-\mathrm{y}=0 & 2x=0 & 2-y=0\\ & x=0 & y=2\\ & (0,0) & (1,2)\\ & & \\ & (y=0) & (x=0)\\ 4x+2y=12 & 4x=12 & 2y=12\\ & x=3 & y=64\\ & (3,0) & (0,6) \end{array}\right]$

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