Find what x could be in this equation

Sorry Tony!. I only have ONE solution for you!.

 

Solve for x:
x/(x - 3) = (2 x)/(x + 6)

Cross multiply:
x (x + 6) = 2 x (x - 3)

Expand out terms of the left hand side:
x^2 + 6 x = 2 x (x - 3)

Expand out terms of the right hand side:
x^2 + 6 x = 2 x^2 - 6 x

Subtract 2 x^2 - 6 x from both sides:
12 x - x^2 = 0

Factor x and constant terms from the left hand side:
-(x (x - 12)) = 0

Multiply both sides by -1:
x (x - 12) = 0

Split into two equations:
x - 12 = 0 or x = 0

Add 12 to both sides:
Answer: |x = 12        or        x = 0

Hmm.  Guest #1 seems to have forgotten the initial 3.

 

3 - x/(x-3) = 2x/(x+6)

 

Multiply by (x-3)(x+6):

 

3(x-3)(x+6) -x(x+6) = 2x(x-3)

 

3(x^2 + 3x - 18) - x^2 - 6x  = 2x^2 - 6x

 

2x^2 + 3x - 54 = 2x^2 - 6x

 

9x = 54

 

x = 6

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