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 $x+7 = 0 => x=-7$ .

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