Question 1103419
{{{x/(x-4) - 4/(3x^2-17x+20) = (2x)/(3x-5)}}}
Factor the 2nd denominator 
{{{x/(x-4) - 4/((3x-5)(x-4) )= (2x)/(3x-5)}}}
multiply by(x-4)(3x-5), cancel the denominators
x(3x-5) - 4 = 2x(x-4)
Distribute
3x^2 - 5x - 4 = 2x^2 - 8x
Combine like terms on the left
3x^2 - 2x^2 - 5x + 8x - 4 = 0
x^2 + 3x - 4 = 0
Factors to
(x+4)(x-1) = 0
Two solutions
x = -4
x = 1
:
Check both values of x ensure they are true and don't result in 0 in the dnominators