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: 31

Answer

The solutions are $x=1$ and $x=\dfrac{3}{4}$

Work Step by Step

$2=\dfrac{3}{2x-1}+\dfrac{-1}{(2x-1)^{2}}$ Multiply the whole equation by $(2x-1)^{2}$: $(2x-1)^{2}\Big[2=\dfrac{3}{2x-1}+\dfrac{-1}{(2x-1)^{2}}\Big]$ $2(2x-1)^{2}=3(2x-1)-1$ Evaluate the indicated operations: $2(4x^{2}-4x+1)=6x-3-1$ $8x^{2}-8x+2=6x-3-1$ Take all terms to the left side and simplify: $8x^{2}-8x+2-6x+3+1=0$ $8x^{2}-14x+6=0$ Divide the whole equation by $2$: $\dfrac{1}{2}\Big(8x^{2}-14x+6=0\Big)$ $4x^{2}-7x+3=0$ Solve by factoring: $(4x-3)(x-1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $4x-3=0$ $4x=3$ $x=\dfrac{3}{4}$ $x-1=0$ $x=1$ The solutions are $x=1$ and $x=\dfrac{3}{4}$
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