We have : 5/(x^2 + 4x) = 3/x - 2/(x + 4)5x2+4x=3x−2x+4.
Let's write all the fractions with the same denominator :
5/(x^2 + 4x) = 3/x - 2/(x + 4)5x2+4x=3x−2x+4
5/(x^2 + 4x) = (3*(x+4))/(x*(x+4)) - (2*x)/((x + 4)*x)5x2+4x=3⋅(x+4)x⋅(x+4)−2⋅x(x+4)⋅x
5/(x^2 + 4x) = (3*(x+4))/(x^2+4x) - (2*x)/(x^2+4x)5x2+4x=3⋅(x+4)x2+4x−2⋅xx2+4x
Now, let's put all the fractions on the left :
5/(x^2 + 4x) = (3x+12)/(x^2+4x) - (2x)/(x^2+4x)5x2+4x=3x+12x2+4x−2xx2+4x
5/(x^2 + 4x) - (3x+12)/(x^2+4x) + (2x)/(x^2+4x) = 05x2+4x−3x+12x2+4x+2xx2+4x=0
(5-(3x+12)+ (2x))/(x^2+4x) = 05−(3x+12)+(2x)x2+4x=0
(5-3x-12+2x)/(x^2+4x) = 05−3x−12+2xx2+4x=0
(-x-7)/(x^2+4x) = 0−x−7x2+4x=0
-(x+7)/(x^2+4x) = 0−x+7x2+4x=0
We can multiply all the equation by -(x^2+4x)−(x2+4x) :
x+7 = 0x+7=0
That equation = 0=0 when x+7 = 0 => x=-7x+7=0⇒x=−7.
