## Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 6 - Systems of Equations and Inequalities - 6-2 Solving Systems Using Substitution - Practice and Problem-Solving Exercises - Page 375: 19

#### Answer

Solution: $(-12,-5)$ (checked below)

#### Work Step by Step

Step 1 Solve one of the equations for one of the variables. Select the first equation as the x has a coefficient of -1. $2=2y-x\qquad$ ... add $x-2$ $x=2y-2$ Step 2 Substitute $2y-2$ for $x$ in the other equation and solve for $y$ $23=5y-4(2y-2)\qquad$ ... (distribute, simplify) $23=5y-8y+8\qquad$ ... add $-8$, simplify $15=-3y\qquad$ ... divide with -3 $y=-5$ Step 3 Substitute $-5$ for $y$ in $x=2y-2$. $x=2(-5)-2$ $x=-12$ Solution: (x,y) = $(-12,-5)$ Check: $\left[\begin{array}{lll} 2\stackrel{?}{=}2(-5)-(-12 & ..... & 23\stackrel{?}{=}5(-5)-4(-12)\\ 2=-10+12 & & 23=-25+48\\ 2=2 & & 23=23\\ & & \end{array}\right]$

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