Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.1 Exercises - Page 10: 1


$x_{1}=-8$ $x_{2}=3$

Work Step by Step

First, change $-2x_{1}-7x_{2}=-5$ to $2x_{1}+7x_{2}=5$ by multiplying both sides by negative 1. $-1*(-2x_{1}-7x_{2})=-1*(-5)$ $2x_{1}+7x_{2}=5$ Next, multiply the first equation by two on both sides to get a common $2x_{1}$ term in both equations, $2*(x_{1}+5x_{2})=2*(7)$ $2x_{1}+10x_{2}=14$ Now you can subtract one equation from the other to get a new equation with ONLY ONE TERM. $2x_{1}+10x_{2}=14$ -$2x_{1}+7x_{2}=5$ The $2x_{1}$ cancels out and you are left with $3x_{2}=9$ Divide both sides by 3 and receive $x_{2}=3$ Now you can plug this into any original equation to receive the answer. $x_{1}+5x_{2}=7$ $x_{1}+5(3)=7$ $x_{1}+15=7$ $x_{1}=7-15=-8$ Check these two values in the other equations to make sure of your answers. $-2x_{1}-7x_{2}=-5$ $-2(-8)-7(3)=-5$ $16-21=-5$ $-5=-5$ These values make sense, and are the answers. Basically, for these types of problems, the idea is to manipulate the equations to find one variable, and plug that variable in to find the other.
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