Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 146: 33


The solutions are $x=3$ and $x=5$

Work Step by Step

$\dfrac{2x-5}{x}=\dfrac{x-2}{3}$ Take $x$ to multiply the numerator of the right side and $3$ to multiply the numerator of the left side: $(2x-5)(3)=(x-2)(x)$ Evaluate the indicated operations: $6x-15=x^{2}-2x$ Take all terms to the right side and simplify: $0=x^{2}-2x-6x+15$ $0=x^{2}-8x+15$ Rearrange: $x^{2}-8x+15=0$ Solve by factoring: $(x-3)(x-5)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-3=0$ $x=3$ $x-5=0$ $x=5$ The solutions are $x=3$ and $x=5$
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