Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.3 - Solving Equations by Using Quadratic Methods - Exercise Set: 7



Work Step by Step

$\dfrac{2}{x}+\dfrac{3}{x-1}=1$ Multiply the whole equation by $x(x-1)$ $x(x-1)\Big(\dfrac{2}{x}+\dfrac{3}{x-1}=1\Big)$ $2(x-1)+3x=x(x-1)$ $2x-2+3x=x^{2}-x$ Take all terms to the right side of the equation and simplify it by combining like terms: $0=x^{2}-x-2x-3x+2$ $x^{2}-6x+2=0$ Solve this equation using the quadratic formula, which is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=1$, $b=-6$ and $c=2$. Substitute the known values into the formula and simplify: $x=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4(1)(2)}}{2(1)}=\dfrac{6\pm\sqrt{36-8}}{2}=...$ $...=\dfrac{6\pm\sqrt{28}}{2}=\dfrac{6\pm2\sqrt{7}}{2}=3\pm\sqrt{7}$ The original equation is not undefined for neither of the values of $x$ found. The final answer is $x=3\pm\sqrt{7}$
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