Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 10: 34

Answer

y= -0.5 x= $\frac{1}{3}$

Work Step by Step

step 1: Find a common factor between one of the variables x or y. In this case we can multiply $\frac{1}{2}$ by 18 to make the x variable equal in both equations. 9x - 4y = 5 $\frac{1}{2} + \frac{1}{3}$ = 0 9x - 4y = 5 $(18) \frac{1}{2} + \frac{1}{3}$ = 0 Step 2: X values cancels out. As y values are opposites you have to take them away. 9x - 4y = 5 9x + 6y = 0 -10y = 5 step 3: rearrange so you get the y value on its own. -y =$ \frac{3}{9}$ y = - $\frac{1}{2}$ Step 4: Use your y answer to solve one of the equations by substituting y by -$\frac{1}{2}$ 9x - 4 (- $\frac{1}{2}$) = 5 Step 5: rearrange the equation so x is left by its own. 9x -4 (-0.50) = 5 9x + 2 = 5 9x = 5 -2 9x = 3 x =$ \frac{3}{9}$ x = $\frac{1}{3}$
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