Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.2 - Linear Equations in One Variable - 1.2 Exercises - Page 87: 40


No solution

Work Step by Step

$x\ne1,-3$ because the denominators cannot be equal to $0$. After multiplying both sides by $(x-1)(x+3):$ $12=3(x+3)+2(x-1)\\12=3x+9+2x-2\\12=5x+7\\5=5x\\x=1$ But $x\ne1$ from earlier. Thus there is no solution.
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