Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 9 - Section 9.3 - Solving Pairs of Linear Equations by Substitution - Exercises - Page 334: 25



Work Step by Step

To solve the following pair of linear equations, we will use the substitution method. $3x+4y=22$ $-5x+2y=-2$ We solve the second equation for $x$. So, $-5x+2y=-2$ $x=\frac{-2-2y}{-5}$ $x=\frac{2+2y}{5}$ First, we substitute the $\frac{2+2y}{5}$ for $x$ in the first equation. $3x+4y=22$ $3(\frac{2+2y}{5})+4y=22$ $\rightarrow$ Simplify $\frac{6}{5}+\frac{6}{5}y+4y=22$ $\frac{26}{5}y=22-\frac{6}{5}$ $\frac{26}{5}y=\frac{104}{5}$ $\rightarrow$ Solving for $y$ $y=\frac{\frac{104}{5}}{\frac{26}{5}}$ $y=\frac{104}{26}$ $y=4$ Now we substitute $4$ for $y$ in any of the equations to solve for $x$. $-5x+2y=-2$ $-5x+2(4)=-2$ $-5x+8=-2$ $-5x=-2-8$ $-5x=-10$ $x=\frac{-10}{-5}$ $x=2$ The apparent solution is $(2,4)$. To check, we substitute $x=2$ and $y=4$ in any of the equations to see if the statement is true or false. If it is true, our solutions are correct. If it is false, the system of equations has no solution. So, $3x+4y=22$ $3(2)+4(4)=22$ $6+16=22$ $22=22$ $\rightarrow$ true Thus, the solution is $(2,4)$.
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