March 1st, 2010

If the equations are separate:
Declare that x =/= -3, -4
Using Fund. Theorem of Algebra, polynomial expansion
(12x+19)/[(x+3)(x+4)] = A/(x+3) + B/(x+4), A = (12*-3+19)/(-3+4) = -17, B = (12*-4 + 19)/(-4+3) = 29
(12x+19)/[(x+3)(x+4)] = -17/(x+3) + 29/(x+4)
Substituting (12x+19)/[(x+3)(x+4)] for its polynomial expansion
3/(x+3) = -17/(x+3) + 29/(x+4) -5/(x+4)
Bringing -17/... to the left, adding 29/... and -5/... together
20/(x+3) = 24/(x+4)
Multiplying both sides to clear denoms
20(x+4) = 24(x+3)
Solving for x
20x + 80 = 24x + 72
8 = 4x
x = 2

2nd equation
Declare that x =/= -4, 0
Clear denoms
5x = (x+4)(x-4)
Expand and solve for 0
x^2 -5x - 16x = 0
Not factorable, use quadratic formula
[-b +or- sqrt(b^2 - 4ac)] / 2a
where a = 1, b = -5, c = -16
x = {-5 + sqrt<[-5]^2 - 4(1)(-16)>} / 2*1 & x = {-5 - sqrt<[-5]^2 - 4(1)(-16)>} / 2*1
x = [-5 + sqrt(89)]/2 & x = [-5 - sqrt(89)]/2
x ~= 7.22 & x ~= -2.22

If x has to satisfy both systems, that will be a small nightmare.

I'll take you up on the nachos and beer, but I'm still taking over the world damnit