Question 75687
I assume you mean: (a few brackets would help)
:
{{{((x+4))/((x^3-3x^2-4x+12))}}} + {{{3/((x^2-5x+6))}}} = {{{(3x)/((x^2-4))}}}
:
All the denominators will factor:
{{{((x+4))/((x-2)(x+2)(x-3)))}}} + {{{3/((x-3)(x-2)))}}} = {{{(3x)/((x-2)(x+2))}}}
:
A common denominator is (x-2)(x+2)(x-3), multiply each term by this and you have:
:
(x+4) + 3(x+2) = 3x(x-3)
:
x + 4 + 3x + 6 = 3x^2 - 9x
:
4x + 10 = 3x^2 - 9x
:
Arrange as a quadratic equation:
3x^2 - 9x - 4x - 10 = 0
3x^2 - 13x - 10 = 0
:
This factors to:
(3x + 2)(x -5) = 0
:
3x = -2
x = -2/3
and
x = +5
:
I substituted 5 for x in the original equation and got equality,
I'll let you check out the x = -2/3 solution