Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by Substitution - Practice - Page 299: 6

Answer

no solutions

Work Step by Step

$\begin{cases} 2x - 3y = 6 \\ -4x + 6y = 12 \end{cases}$ We solve for x from the first equation: $2x-3y=6$ $2x=6+3y$ $x=3+\frac{3}{2}y$ Now that we have a solution for x, we substitute $x=3+\frac{3}{2}y$ in the second equation: $-4x + 6y = 12 $ $-4(3+\frac{3}{2}y) + 6y = 12 $ $-12-6y + 6y = 12 $ $-12 = 12 $ Since this statement is never true, the equation has no solutions.
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