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 - Exercise Set: 29

Answer

x = 2 y = 1

Work Step by Step

5x + 10 y = 20 2x + 6y = 10 to the second equation divide by 3 x + 3y = 5 minu 3y to both sides x + 3y - 3y = 5 - 3y x = 5 - 3y substitude x to the first equation 5 ( 5 - 3y ) + 10 y = 20 25 - 15 y + 10y = 20 25 - 5y = 20 minus 25 to both sides 25 - 5y - 25 = 20 - 25 -5y = -5 divide by -5 both sides $\frac{-5y}{-5}$ = $\frac{-5}{-5}$ y = 1 substitude y = 1 to x = 5 - 3y x = 5 - $3\times1$ x = 5 - 3 x = 2
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