How do you solve # 5 /(x^2 + 4x) = 3 / x - 2/(x + 4)#?

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Answer 1 · 2015-06-02T23:03:24

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#.