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Hi, please help:

\({{x+2}\over x^2-2x}-{{2}\over x-2}=-1\)

 

i need to have X solved...i have no idea...thank you kindly..

 Jul 23, 2018
 #1
avatar+25267 
+1

solve for X
\dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2}=-1

\(\dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2}=-1\)

 

\(\begin{array}{|rcll|} \hline \dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2} &=& -1 \quad | \quad x-2 \ne 0 \Rightarrow x \ne 2 \\\\ && \quad | \quad x^2-2x \ne 0 \Rightarrow x \ne 2 \\\\ \dfrac{(x+2)(x-2)-2(x^2-2x)}{(x^2-2x)(x-2)} &=& -1 \\\\ (x+2)(x-2)-2(x^2-2x) &=& -(x^2-2x)(x-2) \\ x^2-4-2x^2+4x &=& -(x^2-2x)(x-2) \\ -x^2+4x-4 &=& -(x^2-2x)(x-2) \quad | \quad \cdot (-1) \\ x^2-4x+4 &=& (x^2-2x)(x-2) \\ (x-2)(x-2) &=& (x^2-2x)(x-2) \quad | \quad : (x-2) \\ x-2 &=& x^2-2x \\ x^2-2x &=& x-2 \\ x^2-3x+2 &=& 0 \\ (x-1)(x-2) &=& 0 \\\\ x = 1 && x \ne 2 \\ \hline \end{array}\)

 

laugh


 

 Jul 23, 2018
 #3
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0

Heureka,

 

thank you very much, although I fail to understand something:

 

why can't x be equal to 2?

Guest Jul 23, 2018
 #3
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0

Heureka,

 

thank you very much, although I fail to understand something:

 

why can't x be equal to 2?

Guest Jul 23, 2018
 #7
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0

If you look closely at the denominator, if x = 2 then you have: x - 2 =2 - 2 =0!!. You should know by now that you cannot divide by zero, because it is "undefined".

Guest Jul 23, 2018
 #8
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Heureka,

 

no worries, I can see why not....thank you very much!!

Guest Jul 23, 2018
 #8
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0

Heureka,

 

no worries, I can see why not....thank you very much!!

Guest Jul 23, 2018
 #10
avatar+25267 
0

why can't x be equal to 2?

 

Hello Guest,

 

x can't be equal to 2, because then the denominator is zero.

We must not divide by zero.

 

laugh

heureka  Jul 24, 2018
 #2
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0

Solve for x:
(x + 2)/(x^2 - 2 x) - 2/(x - 2) = -1

Bring (x + 2)/(x^2 - 2 x) - 2/(x - 2) together using the common denominator x:
-1/x = -1

Take the reciprocal of both sides:
-x = -1

Multiply both sides by -1:

x = 1

 Jul 23, 2018
 #5
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0

thanx guest,

 

but your answer reads difficultly..

Guest Jul 23, 2018
 #5
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0

thanx guest,

 

but your answer reads difficultly..

Guest Jul 23, 2018

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