Algebra: A Combined Approach (4th Edition)

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

Chapter 4 - Section 4.3 - Solving Systems of Linear Equations by Addition - Practice: 6

Answer

($-\frac{17}{2}, -\frac{3}{2}$)

Work Step by Step

1. Multiply each equation by its LCD. 3($-\frac{x}{3} + y) = 3(\frac{4}{3})$ 2($\frac{x}{2} - \frac{5}{2}y) = 2(-\frac{1}{2})$ Simplifies to: -x + 3y = 4 x - 5y = -1 2. Add the equations and solve for y: -2y = 3 y = $-\frac{3}{2}$ 3. Let y = $-\frac{3}{2}$ in the first equation to find x: $-\frac{x}{3} + (-\frac{3}{2}) = \frac{4}{3}$ $-\frac{x}{3} = \frac{4}{3} + \frac{3}{2}$ $-\frac{x}{3} = \frac{17}{6}$ x = $-\frac{17}{2}$ 4. Check the solution ($-\frac{17}{2}, -\frac{3}{2}$) in the second equation: $\frac{-\frac{17}{2}}{2} - \frac{5}{2}(-\frac{3}{2}) = -\frac{1}{2}$ $\frac{-17}{4} + \frac{15}{4} = -\frac{1}{2}$ $-\frac{2}{4} = -\frac{1}{2}$ $ -\frac{1}{2} = -\frac{1}{2}$ True
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