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: 21

Answer

There is one solution: (2, 1), or x = 2, y = 1.

Work Step by Step

Equation 1: 0.05x - 0.03y = 0.07 Equation 2: 0.07x + 0.02y = 0.16 Let's start by solving for the x variable. To eliminate variable y, we can get both y values to be their lowest common multiple, which is 0.06 for the numbers 0.02 and 0.03. 1: 2(0.05x - 0.03y) = 2(0.07) 0.1x - 0.06y = 0.14 (distributive property) 2: 3(0.07x + 0.02y) = 3(0.16) 0.21x + 0.06y = 0.48 Our new 2 equations contain a +0.06y and -0.06y. We can add the equations to eliminate this variable. Equation 1: 0.1x - 0.06y = 0.14 Equation 2: + 0.21x + 0.06y = 0.48 = .31x + 0y = 0.62 We can eliminate the 0y because it cancels out. This leaves us with 0.31x = 0.62, and dividing both sides by 0.31 gets x = 2 (.62 / .31= 2). To solve for y, plug the x value (2) into either of the original equations. 0.05(2) - 0.03y = 0.07 (plug in) 0.1 - 0.03y = 0.07 (simplify) -0.03y = -0.03 (subtract 0.1 from both sides) y = 1 (divide each side by -0.03)
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