SOLUTION: Please Help Me Solve This Equation: {{{6/(x-1) = 4/(x-2) - 2/(x+1)}}} would the answer be no solution?

Question 32803: Please Help Me Solve This Equation: would the answer be no solution?
Found 2 solutions by longjonsilver, sarah_adam:
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
multiply all terms by (x-1)(x+1)(x-2)

This does not factorise easily, so either complete the square or use the quadratic formula:

These are the 2 precise answers

jon.

Answer by sarah_adam(201) (Show Source): You can put this solution on YOUR website!
6/(x-1) = 4(x+1) - 2(x-2)/ (x-2)(x+1)

6/(x-1) = 4x + 4- 2x - 4/(x-2)(x+1)

6/(x-1) = 4x-2x+4-4/ (x-2)(x+1) Hint (4x-2x = 2x; +4 � 4 = 0)

6/(x-1) = 2x/ (x-2)(x+1)

Now cross multiply:
6(x-2) (x+1) = 2x(x-1)
6(x^2 + x � 2x -2) = 2x^2 -2x
6(x^2 � x -2) =2(x^2 � x)
Now divide the entire equation with 2
6(x^2 � x -2)/2 = 2(x^2 � x)/2

3(x^2 � x -2) = (x^2 � x)
3x^2 � 3x � 6 = x^2 � x
Taking the like terms to same side:
3x^2 � 3x � 6 � x^2 + x = 0
3x^2 � x^2 -3x + x - 6 =0
2x^2 � 2x � 6 = 0
2(x^2 �x � 3) = 0
x^2 � x -3 =0
Solving the equation:
x = -b +sqrt(b^2+ 4*a*c)/2*a or x = -b - sqrt(b^2 - 4*a*c)/2*a
where a = coefficient of x^2
b = coefficient of x
c = constant
so here in this equation a = 1 ; b = -1 ; c = -3
Therfore
x = 1+ sqrt(1+12)/2 ==> x = 1+ sqrt(13)/2
or
x = 1- sqrt(1+12)/2 ==> x = 1-sqrt(13)/2

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