## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 5, Systems of Equations and Inequalities - Section 5.1 - Systems of Linear Equations in Two Variations - 5.1 Exercises: 1

#### Answer

(5,-1) is not a solution. (-1,3) is not a solution. (2,1) is a solution.

#### Work Step by Step

Given a system of linear equations, we check if an ordered pair is a solution by substituting the given x and y value into the system. If both equations produce a true statement, the ordered pair is a solution and represents the intersection of both linear equations. If at least one equations produces a false statement, the ordered pair is not a solution to the system. Consider the following system of equations. $2x+3y=7$ $5x-y=9$ We need to first test if (5, -1) is a solution. To do so we will substitute 5 for x and -1 for y and evaluate each equation. $2(5)+3(-1)=7$ $10-3=7$ $7=7$ True $5(5)-(-1) = 9$ $25+1 = 9$ $26 = 9$ False Because at least one evaluation is false, (5,-1) is not a solution. onsider the following system of equations. $2x+3y=7$ $5x-y=9$ We need to first test if (-1, 3) is a solution. To do so we will substitute -1 for x and 3 for y and evaluate each equation. $2(-1)+3(3)=7$ $-2-9=7$ $7=7$ True $5(-1)-(3) = 9$ $-5-3 = 9$ $-8 = 9$ False Because at least one evaluation is false, (-1,3) is not a solution. We need to first test if (2, 1) is a solution. To do so we will substitute 2 for x and 1 for y and evaluate each equation. $2(2)+3(1)=7$ $4+3=7$ $7=7$ True $5(2)-(1) = 9$ $10-1 = 9$ $9 = 9$ True Because both evaluations are true, (2,1) is a solution.

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