College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.1 - Linear Equations - 1.1 Assess Your Understanding - Page 90: 54

Answer

The equation has no solutions.

Work Step by Step

STEP 1: Restrictions. There must be no zeros in the denominator, so $x\neq\pm 3$ STEP 2: Fractions. The LCM of all denominators is $x^{2}-9=(x+3)(x-3).$ After multiplying both sides with the LCM, $x+4(x-3)=3$ STEP 3: Remove all parentheses and simplify. $x+4x-12=3$ $5x-12=3$ STEP 4: Variables on one side, all other terms on the other... $5x-12=3\qquad/+12$ $5x=15$ STEP 5: Simplify and solve. $5x=15\qquad /\div 5$ $x=3$ In step 1, the number $3$ is excluded as a possible solution. Thus, the equation has no solutions.
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