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Solve for x: \(\displaystyle {{x+2}\over x^2-2x}-{{2}\over x-2}=-1\)

 Jun 28, 2020
 #1
avatar+24342 
+1

First look to see that x cannot equal 2    because denominators would = 0

 

Now change L side by getting common denominator

 

x^2-4    - 2x^2+4x        = -1

 (x^2-2x)(x-2) 

 

the top factors to    (x-2)(-x+2)        then cancel out the x-2 in the top and bottom and you have

 

 

(-x+2)

(x^2-2x)       = -1

 

or     -x+2 = 2x-x^2     

x^2-3x+2

(x-2)(x-1) = 0      `     solve for   x = 1    or  2  (not allowed)

 
 Jun 28, 2020
 #2
avatar+8205 
0

In the fullest details:

\(\dfrac{x + 2}{x^2 - 2x} - \dfrac2{x - 2} = -1\\ \dfrac{x + 2}{x(x - 2)} - \dfrac{2}{x - 2} = -1\\ \dfrac{x + 2}{x(x - 2)} - \dfrac{2\color{blue}x\color{black}}{\color{blue}x\color{black}(x - 2)} = -1\\ \dfrac{(x + 2) - 2x}{x(x - 2)} = -1\\ -x + 2 = -x(x - 2)\\ -x + 2 = -x^2 + 2x\\ x^2 -3x + 2 = 0\\ (x - 1)(x - 2) = 0\\ x = 1 \text{ or } x = 2\text{ (rejected)}\)

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 Jun 28, 2020
edited by MaxWong  Jun 28, 2020

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