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: 30


The solutions are $x=-\dfrac{1}{3}$ and $x=\dfrac{7}{2}$

Work Step by Step

$\dfrac{7}{x^{2}}+\dfrac{19}{x}=6$ Multiply the whole equation by $x^{2}$: $x^{2}\Big(\dfrac{7}{x^{2}}+\dfrac{19}{x}=6\Big)$ $7+19x=6x^{2}$ Take all terms to the right side of the equation and rearrange: $0=6x^{2}-19x-7$ $6x^{2}-19x-7=0$ Solve by factoring: $(3x+1)(2x-7)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $3x+1=0$ $3x=-1$ $x=-\dfrac{1}{3}$ $2x-7=0$ $2x=7$ $x=\dfrac{7}{2}$ The solutions are $x=-\dfrac{1}{3}$ and $x=\dfrac{7}{2}$
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