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: 16

Answer

The solution is $(-17,5)$.

Work Step by Step

The given system of equations is $-2x-5y=9$ ...... (1) $3x+11y=4$ ...... (2) Multiply each side of equation (1) by $3$. $3(-2x-5y)=3(9)$ Simplify. $-6x-15y=27$ ...... (3) Multiply each side of equation (2) by $2$. $2(3x+11y)=2(4)$ Simplify. $6x+22y=8$ ...... (4) Add equation (3) and (4). $\Rightarrow -6x-15y+6x+22y=27+8$ Add like terms. $\Rightarrow 7y=35$ Divide each side by $7$. $\Rightarrow \frac{7y}{7}=\frac{35}{7}$ Simplify. $\Rightarrow y=5$ Substitute $5$ for $y$ in equation (2). $\Rightarrow 3x+11(5)=4$ Simplify. $\Rightarrow 3x+55=4$ Subtract $55$ from each side. $\Rightarrow 3x+55-55=4-55$ Simplify. $\Rightarrow 3x=-51$ Divide each side by $3$. $\Rightarrow \frac{3x}{3}=\frac{-51}{3}$ Simplify. $\Rightarrow x=-17$ Check $(x,y)=(-17,5)$ Equation (1): $\Rightarrow -2x-5y=9$ $\Rightarrow -2(-17)-5(5)=9$ $\Rightarrow 34-25=9$ $\Rightarrow 9=9$ True. Equation (2): $\Rightarrow 3x+11y=4$ $\Rightarrow 3(-17)+11(5)=4$ $\Rightarrow -51+55=4$ $\Rightarrow 4=4$ True. Hence, the solution is $(-17,5)$.
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