Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 5 - Solving Systems of Linear Equations - 5.2 - Solving Systems of Linear Equations by Substitution - Monitoring Progress - Page 242: 3

Answer

$(-1,1)$

Work Step by Step

The given system of equations is $\Rightarrow x=6y-7$ ...... (1) $\Rightarrow 4x+y=-3$ ...... (2) Substitute $6y-7$ for $x$ in equation (2). $\Rightarrow 4(6y-7)+y=-3$ Use distributive property. $\Rightarrow 24y-28+y=-3$ Add like terms. $\Rightarrow 25y-28=-3$ Add $28$ to each side. $\Rightarrow 25y-28+28=-3+28$ Simplify. $\Rightarrow 25y=25$ Divide each side by $25$. $\Rightarrow \frac{25y}{25}=\frac{25}{25}$ Simplify. $\Rightarrow y=1$ Substitute $1$ for $y$ in equation (1). $\Rightarrow x=6(1)-7$ Simplify. $\Rightarrow x=6-7$ $\Rightarrow x=-1$ Check Equation (1) $\Rightarrow x=6y-7$ $\Rightarrow -1=6(1)-7$ $\Rightarrow -1=6-7$ $\Rightarrow -1=-1$ True. Check Equation (2) $\Rightarrow 4x+y=-3$ $\Rightarrow 4(-1)+1=-3$ $\Rightarrow -4+1=-3$ $\Rightarrow -3=-3$ True. Hence, the solution is $(-1,1)$.
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