Linear Algebra and Its Applications (5th Edition)

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

Chapter 1 - Linear Equations in Linear Algebra - 1.1 Exercises - Page 10: 2


$x_{1}=12$ $x_{2}=-7$

Work Step by Step

This problem requires alot of manipulation of the variables. To start, divide the first equation by two so that you can begin to eliminate $x_{1}$. $\frac{(2x_{1}+4x_{2})}{2}=\frac{-4}{2}$ $x_{1}+2x_{2}=-2$ Now you can freely manipulate $x_{1}$, as in you can now set the constant of $x_{1}$ to whatever you would like. For example, you can multiply the equation by 10 to change $x_{1}$ to $10x_{1}$. We will use this to eliminate $x_{1}$ and isolate $x_{2}$. We know that in the second equation, $x_{1}$ is multiplied by $5$, so if we multiply the first equation by 5, we can eliminate $x_{1}$. $5*(x_{1}+2x_{2})=5*(-2)$ $5x_{1}+10x_{2}=-10$ And now subtract one equation from the other. $(5x_{1}+7x_{2}=11)$ $-(5x_{1}+10x_{2}=-10)$ And you will get $-3x_{2}=21$, and you can solve for $x_{2}$. $-3x_{2}=21$ $x_{2}=\frac{21}{-3}=-7$ Now you can find $x_{1}$ through substitution. $2x_{1}+4x_{2}=-4$ $2x_{1}+4(-7)=-4$ (Substitution) $2x_{1}-28=-4$ $2x_{1}=-4+28=24$ (Getting $x_{1}$ alone) $x_{1}=\frac{24}{2}=12$ (Solving for $x_{1}$) Lastly, check by substituting into the other equation. $5x_{1}+7x_{2}=11$ $5(12)+7(-7)=11$ (Substitution) $60-49=11$ $11=11$ The values work and are therefore the answers. Same principle is used in this one, only this one had a bit more manipulation.
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