Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises - Page R-16: 33


The solution is $x=\dfrac{2}{3}$

Work Step by Step

$\dfrac{1}{x-2}-\dfrac{3x}{x-1}=\dfrac{2x+1}{x^{2}-3x+2}$ Factor the denominator of the fraction on the right side of the equation: $\dfrac{1}{x-2}-\dfrac{3x}{x-1}=\dfrac{2x+1}{(x-2)(x-1)}$ Multiply the whole equation by $(x-2)(x-1)$: $(x-2)(x-1)\Big[\dfrac{1}{x-2}-\dfrac{3x}{x-1}=\dfrac{2x+1}{(x-2)(x-1)}\Big]$ $x-1-3x(x-2)=2x+1$ $x-1-3x^{2}+6x=2x+1$ Take all terms to the right side and simplify: $0=3x^{2}+2x-6x-x+1+1$ $0=3x^{2}-5x+2$ Rearrange: $3x^{2}-5x+2=0$ Solve by factoring: $(3x-2)(x-1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $3x-2=0$ $3x=2$ $x=\dfrac{2}{3}$ $x-1=0$ $x=1$ The original equation is undefined for $x=1$, so the solution is $x=\dfrac{2}{3}$
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