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: 78


We have to choose $$k=-\frac{2}{3}.$$

Work Step by Step

Step 1: Make the right sides of the equations to be equal. To do this multiply the first equation by $-3$: \begin{align*} -3kx-3y=&-12\\ 2x-3y=&-12 \end{align*} Step 2: Find such $k$ for which the left sides of the two equations will be completely identical (same coefficients multiplying $x$ and $y$). In this way we will have only one independent equation and that means infinitely many solutions. We need that $-3k = 2$ which means that $$k = -\frac{2}{3}.$$ Now we really have \begin{align*} 2x-3y=&-12\\ 2x-3y=&-12 \end{align*} which means we can drop one equation and the system will have infinitely many solutions.
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