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: 72


The solution to this system is: $$x=-\frac{34}{25},\quad y=\frac{51}{25}.$$

Work Step by Step

As instructed in the problem we will set $A=1/x$ and $B=1/y$. Then we get a system of LINEAR equations in $A$ and $B$: \begin{align*} 2A+3B=&0\\ 3A-4B=&-\frac{25}{6}. \end{align*} Now let's solve this system: Step 1: Express $B$ in terms of $A$ from the first equation: $$2A+3B=0\Rightarrow 3B=-2A$$ and that gives $$B=-\frac{2}{3}A.$$ Step 2: Put this into the second equation and solve for $A$: $$3A-4\times(-\frac{2}{3}A) = -\frac{25}{6}.$$ This further gives $$3A+\frac{8}{3}A=-\frac{25}{6}\Rightarrow \frac{9}{3}A+\frac{8}{3}A=-\frac{25}{6}\Rightarrow \frac{17}{3}A=-\frac{25}{6}.$$ This now yields $$A=-\frac{25}{6}\times\frac{3}{17} = -\frac{25}{34}.$$ Step 3: Use this calculated value to solve for $B$: $$B=-\frac{2}{3}A=-\frac{2}{3}\times \left(-\frac{25}{34}\right) = \frac{25}{51}.$$ Now we have to solve for $x$ and $y$ and to do this we have to remember that: $$A=\frac{1}{x}$$ and this means that $$x=\frac{1}{A}=-\frac{34}{25}.$$ Also the fact that $$B=\frac{1}{y}$$ means that $$y=\frac{1}{B} = \frac{51}{25}.$$
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