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: 15

Answer

One solution: (4,1) (x = 4, y = 1)

Work Step by Step

Equation 1: 3x - 5y = 7 Equation 2: 2x + y = 9 Solving this by elimination is the easiest route. It is simplest to multiply equation 2 by 5; this way, we will have a -5y and +5y to cancel out in both equations. Our new equation 2 is: 5(2x + y) = 5(9) 10x + 5y = 45 (distributive property, multiplication) If we add equation 1 and our new equation 2, we get... Equation 1: 3x - 5y = 7 Equation 2: + 10x + 5y = 45 13x + 0y = 52 The 0y can be eliminated, so we are left with 13x = 52. Dividing both sides by 13 gets x by itself, which equals 4 (52 divided by 13). To solve for the y-coordinate, plug x into either original equation and solve. 2(4) + y = 9 (plug x-value into original equation) 8 + y = 9 (multiply) y = 1 (solve for y by isolating and subtracting 8 from each side) There is 1 solution for the equations: (4,1), or x = 4, y = 1.
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