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