Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 6 - Systems of Equations and Inequalities - 6-2 Solving Systems Using Substitution - Practice and Problem-Solving Exercises - Page 375: 22

Answer

Solution: $(1,3)$ (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. $ 4y-x=5+2y\qquad$ ... add $x-5-2y$ $4y-2y-5=x$ $x=2y-5$ Step 2 Substitute $2y-5$ for $x$ in the other equation and solve for $y$ $ 3(2y-5)+7y=24\qquad$ ... (distribute, simplify) $ 6y-15+7y=24\qquad$ ... add $15$, simplify $ 13y=39\qquad$ ... divide with $13$ $y=3$ Step 3 Substitute $3$ for $y$ in $x=2y-5$. $x=2(3)-5$ $x=1$ Solution: (x,y) = $(1,3)$ Check: $\left[\begin{array}{lll} 4(3)-1\stackrel{?}{=}5+2(3) & ..... & 3(1)+7(3)\stackrel{?}{=}24 \\ 12-1=5+6 & & 3+21=24\\ 11=11 & & 24=24\\ & & \end{array}\right]$
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