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delta23
KarishmaB Bunuel, please help

Easiest way to do it is to plug answer choices in and see which one works. Only E works.
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Would there be a way to solve this in under 2 minutes?

I tried simplifying the equation to reach the answer and it took me approx. 7 minutes to do so.

Thanks!
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Would there be a way to solve this in under 2 minutes?

I tried simplifying the equation to reach the answer and it took me approx. 7 minutes to do so.

Thanks!

OA: E

\(\frac{x}{2(x-1)}+\frac{2}{x+1}=\frac{2x}{1+x}+\frac{1}{2(1-x)}\)

Multiplying both sides by \(2(x-1)(x+1)\),we get

\(x(x+1)+4(x-1)=4x(x-1)-(x+1)\)

\(x^2+x+4x-4=4x^2-4x-x-1\)

\(3x^2 -10x+3=0\)
\(3x^2 -9x-1x+3=0\)
\(3x(x-3)-1(x-3)=0\)
\((3x-1)(x-3)=0\)
\(x=\frac{1}{3}\)(Rejected as \(x\) should be an integer), \(3\)
\(x=3\)
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WIKI18
Would there be a way to solve this in under 2 minutes?

I tried simplifying the equation to reach the answer and it took me approx. 7 minutes to do so.

Thanks!

OA: E

\(\frac{x}{2(x-1)}+\frac{2}{x+1}=\frac{2x}{1+x}+\frac{1}{2(1-x)}\)

Multiplying both sides by \(2(x-1)(x+1)\),we get

\(x(x+1)+4(x-1)=4x(x-1)-(x+1)\)

\(x^2+x+4x-4=4x^2-4x-x-1\)

\(3x^2 -10x+3=0\)
\(3x^2 -9x-1x+3=0\)
\(3x(x-3)-1(x-3)=0\)
\((3x-1)(x-3)=0\)
\(x=\frac{1}{3}\)(Rejected as \(x\) should be an integer), \(3\)
\(x=3\)

When I multiply the last term i.e. \(\frac{1}{2(1-x)}\) by \(2(x-1)(x+1)\) I get \(\frac{(x-1)(x+1)}{(1-x)}\)

Can you please explain how you reached\((x+1)\)?
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WIKI18
Would there be a way to solve this in under 2 minutes?

I tried simplifying the equation to reach the answer and it took me approx. 7 minutes to do so.

Thanks!

OA: E

\(\frac{x}{2(x-1)}+\frac{2}{x+1}=\frac{2x}{1+x}+\frac{1}{2(1-x)}\)

Multiplying both sides by \(2(x-1)(x+1)\),we get

\(x(x+1)+4(x-1)=4x(x-1)-(x+1)\)

\(x^2+x+4x-4=4x^2-4x-x-1\)

\(3x^2 -10x+3=0\)
\(3x^2 -9x-1x+3=0\)
\(3x(x-3)-1(x-3)=0\)
\((3x-1)(x-3)=0\)
\(x=\frac{1}{3}\)(Rejected as \(x\) should be an integer), \(3\)
\(x=3\)

When I multiply the last term i.e. \(\frac{1}{2(1-x)}\) by \(2(x-1)(x+1)\) I get \(\frac{(x-1)(x+1)}{(1-x)}\)

Can you please explain how you reached\((x+1)\)?
WIKI18
\(2(x-1)(x+1)\)*\(\frac{1}{2(1-x)}\)=\(2(x-1)(x+1)\)*\(\frac{1}{-2(x-1)}\) =\((x+1)\)*\(-1=-(x+1)\)
Let me know if further explaination is required.
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Princ


\(2(x-1)(x+1)\)*\(\frac{1}{2(1-x)}\)=\(2(x-1)(x+1)\)*\(\frac{1}{-2(x-1)}\) =\((x+1)\)*\(-1=-(x+1)\)
Let me know if further explaination is required.

Got it now. Thanks!
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Here is an easier approach:
  • \(\frac{x}{2x-2}­\)­ + \(\frac{2}{x+1}­\)­ = \(\frac{2x}{x+1}­\)­ + \(\frac{1}{2-2x}­\)­
  • \(\frac{x}{2x-2}­\)­ - \(\frac{1}{2-2x}­\)­ = \(\frac{2x}{x+1}­\)­ - \(\frac{2}{x+1}­\)­
  • \(\frac{x}{2x-2}­\)­ - \((\frac{-1}{2x-2}­)\)­ = \(\frac{2x - 2}{x+1}­\)
  • \(\frac{x+1}{2x-2}­\)­ = \(\frac{2x - 2}{x+1}­\)
  • \((x+1)^2 = (2x-2)^2\)
solve the quadratic and you get x = 3 or 1/3
 ­
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