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 - 5.4 - Solving Special Systems of Linear Equations - Exercises - Page 257: 16

Answer

The solution is $(3,7)$.

Work Step by Step

The given system of equations is $3x-2y=-5$ ...... (1) $4x+5y=47$ ...... (2) Multiply equation (1) by $5$. $5(3x-2y)=5(-5)$ Use distributive property. $15x-10y=-25$ ...... (3) Multiply equation (2) by $2$. $2(4x+5y)=2(47)$ Use distributive property. $8x+10y=94$ ...... (4) Add equation (3) and (4). $\Rightarrow 15x-10y+8x+10y=-25+94$ Add like terms. $\Rightarrow 23x=69$ Divide each side by $23$. $\Rightarrow \frac{23x}{23}=\frac{69}{23}$ Simplify. $\Rightarrow x=3$ Substitute $3$ for $x$ in equation (2). $\Rightarrow 4(3)+5y=47$ Simplify. $\Rightarrow 12+5y=47$ Subtract $12$ from each side. $\Rightarrow 12+5y-12=47-12$ Simplify. $\Rightarrow 5y=35$ Divide each side by $5$. $\Rightarrow \frac{5y}{5}=\frac{35}{5}$ Simplify. $\Rightarrow y=7$ Hence, the solution is $(3,7)$.
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