## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 10: 44

#### Answer

$x_1$=-1 $x_2$=-1 (-1,-1)

#### Work Step by Step

$\frac{x_1+4}{3}+\frac{x_2+1}{2}=1$ $3x_1-x_2=-2$ Expres $x_2$ in terms of $x_1$. $x_2=2+3x_1$ Substitute for $x_2$ in the first equation. $\frac{x_1+4}{3}+\frac{(2+3x_1)+1}{2}=1$ $\rightarrow$ Simplify. $\frac{x_1+4}{3}+\frac{3+3x_1}{2}=1$ $\rightarrow$ Multiply by 6 to remove fractions. $2(x_1+4)+3(3+3x_1)=6$ $\rightarrow$ Simplify. $2x_1+8+9+9x_1=6$ $11x_1=-11$ $x_1=-1$ Substitute for $x_1$ and solve for $x_2$. $3x_1-x_2=-2$ $3(-1)-x_2=-2$ $x_2=-1$

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