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: 40

Answer

$x=-1$ $y=-\frac{1}{2}$ (-1,$-\frac{1}{2}$)

Work Step by Step

$\frac{3}{4}x-\frac{y}{2}=-\frac{1}{2}$ $x+y=-\frac{3}{2}$ Remove fractions. Multiply both sides of the first equation by 4. $4(\frac{3}{4}x-\frac{y}{2})=4(-\frac{1}{2})\longrightarrow$ Simplify. Apply the distributive property. $3x-2y=-2$ Multiply both sides of the second equation by 2. $2(x+y)=2(-\frac{3}{2})\longrightarrow$ Simplify. Apply the distributive property. $2x+2y=-3$ Add the two new equations together to eliminate y. $3x-2y=-2$ $\underline{2x+2y=-3}$ $5x\ \ \ \ \ \ \ \ \ =-5$ Solve for x. $5x=-5\longrightarrow$ Divide both sides by 5. $5x\div5=-5\div5\longrightarrow$ Simplify. $x=-1$ Substitute for x in any equation to solve for y. $-1+y=-\frac{3}{2}\longrightarrow$ Add 1 to both sides. $-1+y+1=-\frac{3}{2}+1\longrightarrow$ Simplify. $y=-\frac{1}{2}$
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