Algebra: A Combined Approach (4th Edition)

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

Chapter 4 - Review - Page 328: 28

Answer

$$x=\frac{-3}{2}$$ $$y=\frac{15}{2}$$

Work Step by Step

equation 1 $$10x + 2y = 0$$ equation 2 $$3x + 5y = 33$$ Multiply equation 1 by $-3$ and equation 2 by $10$: equation 1: $$[10x + 2y = 0]\cdot-3$$ $$-30x -6y = 0$$ equation 2: $$[3x + 5y = 33]\cdot10$$ $$30x +50y = 330$$ Add both equations: $$-30x -6y = 0$$ $$30x +50y = 330$$ $$44y=330$$ $$y=\frac{330}{44}$$ $$y=\frac{15}{2}$$ Substitute this value of $x$ to equation 1: $$-30x -6y = 0$$ $$-30x -6(\frac{15}{2}) = 0$$ $$-30x -45 = 0$$ $$-30x=45$$ $$x=\frac{45}{-30}$$ $$x=\frac{-3}{2}$$
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