Question

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

Answer 1

`x = -3/4 + sqrt(129)/4 color(white)(xx)` and `color(white)(xx)x = -3/4 - sqrt(129)/4`

Explanation:

First of all, please be aware that `x != 0` and `x != -3` must hold so that the equation is defined.

Now, with these restrictions in mind, let's start solving. :)

First of all, we need to get rid of the denominators. To do so, multiply both sides with `x * (x+3)`:

`color(white)(xxxss) 5/x - 2 = 2 / (x+3)`

`<=> color(white)(xx) 5 ( x + 3 ) - 2 x (x + 3) = 2 x`
`<=> color(white)(xx) 5x + 15 - 2x^2 - 6x = 2x`

... add `-2x` on both sides and simplify ...

`<=> color(white)(xx) 15 - 3x - 2x^2 = 0`

This is a quadratic equation which can be solved for example with the quadratic formula which states that the solution of `ax^2 + bx + c = 0` is

`x = (-b +- sqrt(b^2-4ac))/(2a)`

In our case, `a = -2`, `b = -3` and `c = 15`.

This leads us to

`x = (3 +- sqrt(9 - 4 * (-2)*15))/-4 = (3 +- sqrt(129))/-4`

The two solutions are

`x = -3/4 + sqrt(129)/4 color(white)(xx)` and `color(white)(xx)x = -3/4 - sqrt(129)/4`