Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 12: 71


The solution to this system is $$x=3,\quad y=-4.$$

Work Step by Step

As instructed in the problem here we will let $A=1/x$ and $B=1/y$. Applying this to our system we get \begin{align*} 12A-12B =&7\\ 3A+4B=&0 \end{align*} Let's solve this system for $A$ and $B$: Step 1: Express $B$ in terms of $A$ from the second equation: $$4B=-3A\Rightarrow B=-\frac{3}{4}A.$$ Step 2: Put this into the first equation and solve for $A$ $$12A-12\times(-\frac{3}{4}A) = 7.$$ This gives $$12A+9A=7\Rightarrow 21A = 7$$ so we have $$A=\frac{7}{21} = \frac{1}{3}.$$ Step 3: Now we use this calculated value to solve for $B$: $$B=-\frac{3}{4}A=-\frac{3}{4}\times\frac{1}{3} = -\frac{1}{4}.$$ Now we have to find $x$ and $y$. Since $$A=\frac{1}{x} = \frac{1}{3}$$ we have that $$x=3.$$ Since $$B=\frac{1}{y} = -\frac{1}{4}$$ we have that $$y=-4.$$
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