Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises: 66

Answer

$x=1\pm\dfrac{\sqrt{6}}{3}$

Work Step by Step

$\dfrac{1}{x}+\dfrac{2}{x-1}=3$ Multiply the whole equation by $x(x-1)$: $x(x-1)\Big(\dfrac{1}{x}+\dfrac{2}{x-1}=3\Big)$ $(x-1)+(2x)=3x(x-1)$ $x-1+2x=3x^{2}-3x$ Take all terms to the right side: $0=3x^{2}-3x-2x-x+1$ $3x^{2}-6x+1=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=3$, $b=-6$ and $c=1$. Substitute the known values into the formula and simplify: $x=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4(3)(1)}}{2(3)}=\dfrac{6\pm\sqrt{36-12}}{6}=...$ $...=\dfrac{6\pm\sqrt{24}}{6}=\dfrac{6\pm2\sqrt{6}}{6}=\dfrac{6}{6}\pm\dfrac{2\sqrt{6}}{6}=1\pm\dfrac{\sqrt{6}}{3}$
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