Question

Solve the equation:

Answer 1

\[\frac{ 3x }{ x-5 }=\frac{ 2 }{ x+1}+\frac{ 2x^2+40 }{ x^2-4x-5 }\]

Answer 2

I know that you have to factor or whatever the x^2-4x-5

Answer 3

so it would look like this \[\frac{ 2(x^2+20) }{ (x-5)(x+4)}\]

Answer 4

x^2 -4x -5 = (x-5)(x+1)

Answer 5

\[\frac{3x}{x-5} = \frac{2(x-5) + 2x^2 + 40}{(x-5)(x+1)}\]
so
\[3x(x+1) = 2(x-5) +2x^2 +40\]
combine terms
set equal to zero
solve quadratic using factoring or quadratic formula

Answer 6

x=6, -5? @dumbcow

Answer 7

check your signs ... i get -6,5

but x can't equal 5 because it makes denominator 0 in original equation

Answer 8

so its just -6

Answer 9

@dumbcow thank you for helping me out with this one.

Answer 10

yw