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: 92

Answer

The solution is $x=300$ and $y=315$. This mean that the two graphs intersect. Their slopes are so similar that on the scale presented on the graph they seem almost parallel. To see that they are not parallel, the range of $x$ should be increased.

Work Step by Step

We will first solve this system: Step 1: Express $y$ in terms of $x$ from the first equation: $$20y=21x\Rightarrow y=\frac{21}{20}x$$ Step 2: Put this into the second equation and find $x$: $$13x-12\times\frac{21}{20}x=120$$ which becomes $$13x-\frac{63}{5}x = 120\Rightarrow \frac{2}{5}x=120$$ and this gives $$x=120\times\frac{5}{2} = 300.$$ Step 3: Use this calculated value for $x$ to find $y$: $$y=\frac{21}{20}\times 300 = 315.$$ The solution is $x=300$ and $y=315$. This mean that the two graphs intersect. Their slopes are so similar that on the scale presented on the graph they seem almost parallel. To see that they are not parallel, the range of $x$ should be increased.
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