Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 6 - Systems of Equations and Inequalities - 6-3 Solving Systems Using Elimination - Practice and Problem-Solving Exercises - Page 379: 35



Work Step by Step

Our two equations are: 1. y=(2/3)x+1 2. 2x+3y=27 First, lets multiply the first equation by 3. We get 1. 3y=2x+3 2. 2x+3y=27 Since we have a 3y in the second equation, we can substitute the first equation into the second equation for 3y. We get the equation: 2x+(2x+3)=27 Simplify this further: 4x+3=27 Subtract 3 and divide by 4 on both sides. Then, we get the value of x: x=6. Now that we have the x value, lets find the y value by substituting 6 for x in the first equation. y=(2/3)(6)+1 Simplify further: y=5 Therefore our answer to the system of equation is (6,5)
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