Algebra: A Combined Approach (4th Edition)

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

Chapter 4 - Test - Page 330: 10

Answer

$$x=-\frac{1}{2}$$ $$y=2$$

Work Step by Step

Equation 1: $\frac{1}{2}x+ 2y =\frac{15}{4}$ Equation 2: $4x = -y$ Multiply equation 2 by $-1$: $$[4x = -y]\cdot-1$$ $$-4x = y$$ Substitute this equation to equation 1: $$\frac{1}{2}x+ 2y =\frac{15}{4}$$ $$\frac{1}{2}x+ 2(-4x) =\frac{15}{4}$$ $$\frac{1}{2}x-8x =\frac{15}{4}$$ $$\frac{1}{2}x-\frac{16}{2}x =\frac{15}{4}$$ $$-\frac{15}{2}x =\frac{15}{4}$$ Divide both sides by $-\frac{15}{2}$: $$\frac{-\frac{15}{2}x}{-\frac{15}{2}} =\frac{\frac{15}{4}}{-\frac{15}{2}}$$ $$x=\frac{15}{4}\cdot-\frac{2}{15}$$ $$x=-\frac{1}{2}$$ Substitute this value of $x$ to equation 2: $$4x = -y$$ $$4(-\frac{1}{2})= -y$$ $$-2=-y$$ $$y=2$$ Use equation 1 to check: $$\frac{1}{2}(-\frac{1}{2})+ 2(2) =\frac{15}{4}$$ $$-\frac{1}{4}+ 4 =\frac{15}{4}$$ $$-\frac{1}{4}+ \frac{16}{4} =\frac{15}{4}$$ $$\frac{15}{4}=\frac{15}{4}$$
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