Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Section 4.1 - Solving Systems of Linear Equations in Two Variables - Exercise Set - Page 212: 25

Answer

x=1 y=-2 (1,-2)

Work Step by Step

$5x+2y=1$ $x-3y=7$ Equate the coefficients by multiplying the second equation by 5. $x-3y=7\longrightarrow$ Multiply both sides of the equation by 5. $5(x-3y)=5(7)\longrightarrow$ Simplify. Apply the distributive property. $5x-15y=35$ Subtract the new equation from the first equation to eliminate x and solve for y. $\ \ \ \ \ 5x+2y=1$ $\underline{-(5x-15y=35)}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ 17y=-34$ $17y=-34\longrightarrow$ Divide both sides by 17. $17y\div17=-34\div17\longrightarrow$ Simplify. $y=-2$ Substitute for y and solve for x. $x-3(-2)=7\longrightarrow$ Simplify. $x-(-6)=7$ $x+6=7\longrightarrow$ Subtract 6 from both sides. $x+6-6=7-6\longrightarrow$ Simplify. $x=1$
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