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 - Exercises - Page 251: 12

Answer

The solution is $(2,1)$.

Work Step by Step

The given system of equations is $8x-5y=11$ ...... (1) $4x-3y=5$ ...... (2) Multiply each side of equation (2) by $-2$. $-2(4x-3y)=-2(5)$ Simplify. $-8x+6y=-10$ ...... (3) Add equation (1) and (3). $\Rightarrow 8x-5y-8x+6y=11-10$ Add like terms. $\Rightarrow y=1$ Substitute $1$ for $y$ in equation (1). $\Rightarrow 8x-5(1)=11$ Simplify. $\Rightarrow 8x-5=11$ Add $5$ to each side. $\Rightarrow 8x-5+5=11+5$ Simplify. $\Rightarrow 8x=16$ Divide each side by $8$. $\Rightarrow \frac{8x}{8}=\frac{16}{8}$ Simplify. $\Rightarrow x=2$ Check $(x,y)=(2,1)$ Equation (1): $\Rightarrow 8x-5y=11$ $\Rightarrow 8(2)-5(1)=11$ $\Rightarrow 16-5=11$ $\Rightarrow 11=11$ True. Equation (2): $\Rightarrow 4x-3y=5$ $\Rightarrow 4(2)-3(1)=5$ $\Rightarrow 8-3=5$ $\Rightarrow 5=5$ True. Hence, the solution is $(2,1)$.
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