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: 14

Answer

The solution is $(19,16)$.

Work Step by Step

The given system of equations is $10x-9y=46$ ...... (1) $-2x+3y=10$ ...... (2) Multiply each side of equation (2) by $3$. $3(-2x+3y)=3(10)$ Simplify. $-6x+9y=30$ ...... (3) Add equation (1) and (3). $\Rightarrow 10x-9y-6x+9y=46+30$ Add like terms. $\Rightarrow 4x=76$ Divide each side by $4$. $\Rightarrow \frac{4x}{4}=\frac{76}{4}$ Simplify. $\Rightarrow x=19$ Substitute $19$ for $x$ in equation (2). $\Rightarrow -2(19)+3y=10$ Simplify. $\Rightarrow -38+3y=10$ Add $38$ to each side. $\Rightarrow -38+3y+38=10+38$ Simplify. $\Rightarrow 3y=48$ Divide each side by $3$. $\Rightarrow \frac{3y}{3}=\frac{48}{3}$ Simplify. $\Rightarrow y=16$ Check $(x,y)=(19,16)$ Equation (1): $\Rightarrow 10x-9y=46$ $\Rightarrow 10(19)-9(16)=46$ $\Rightarrow 190-144=46$ $\Rightarrow 46=46$ True. Equation (2): $\Rightarrow -2x+3y=10$ $\Rightarrow -2(19)+3(16)=10$ $\Rightarrow -38+48=10$ $\Rightarrow 10=10$ True. Hence, the solution is $(19,16)$.
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