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 - 5.3 - Solving Systems of Linear Equations by Elimination - Monitoring Progress - Page 249: 3

Answer

The solution is $(2,5)$.

Work Step by Step

The given system of equations is $x+4y=22$ ...... (1) $4x+y=13$ ...... (2) Multiply equation (2) by $-4$. $-4(4x+y)=-4(13)$ Use distributive property. $-16x-4y=-52$ ...... (3) Add equation (1) and (3). $\Rightarrow x+4y-16x-4y=22-52$ Add like terms. $\Rightarrow -15x=-30$ Divide each side by $-15$. $\Rightarrow \frac{-15x}{-15}=\frac{-30}{-15}$ Simplify. $\Rightarrow x=2$ Substitute $2$ for $x$ equation (1). $\Rightarrow 2+4y=22$ Subtract $2$ from each side. $\Rightarrow 2+4y-2=22-2$ Simplify. $\Rightarrow 4y=20$ Divide each side by $4$. $\Rightarrow \frac{4y}{4}=\frac{20}{4}$ Simplify. $\Rightarrow y=5$ Check $(x,y)=(2,5)$ Equation (1): $\Rightarrow x+4y=22$ $\Rightarrow 2+4(5)=22$ $\Rightarrow 2+20=22$ $\Rightarrow 22=22$ True. Equation (2): $\Rightarrow 4x+y=13$ $\Rightarrow 4(2)+5=13$ $\Rightarrow 8+5=13$ $\Rightarrow 13=13$ True. Hence, the solution is $(2,5)$.
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