Question

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

Answer 1

`x_(1,2) = (-3 +- sqrt(129))/4`

Explanation:

Your equation looks like this

`5/x - 2 = 2/(x+3)`

Right from the start, you know that `x` cannot take a value that would make the two denominators equal to zero. This means that you need

`x != 0" "` and `" " x + 3 !=0 implies x != -3`

With this in mind, use the common denominator `x * (x+3)` to get rid of the denominators. Multiply the first fraction by `1 = (x+3)/(x+3)`, multiply `2` by `1 = (x(x+3))/(x(x+3))`, and the second fraction by `1 = x/x` to get

`5/x * (x+3)/(x+3) - 2 * (x(x+3))/(x(x+3)) = 2/(x+3) * x/x`

This is equivalent to

`(5 * (x+3))/(x(x+3)) - (2x(x+3))/(x(x+3)) = (2x)/(x(x+3))`

You can thus say that

`5(x+3) - 2x(x+3) = 2x`

Expand the parantheses to get

`5x + 15 - 2x^2 - 6x = 2x`

Rearrange by getting all the terms on one side of the equation

`2x^2 + 3x - 15 = 0`

Use the quadratic formula to find the two solutions

`x_(1,2) = (-3 +- sqrt(3^2 - 4 * 2 * (-15)))/(2 * 2)`

`x_(1,2) = (-3 +- sqrt(129))/4`

The two solutions to the original equation will thus be

`x_1 = (-3 - sqrt(129))/4" "` and `" "x_2 = (-3 + sqrt(129))/4`

Do a quick check to make sure that the calculations are correct - I'll do them for `x = (-3 - sqrt(129))/4`

`5/((-3 - sqrt(129))/4) - 2 = 2/( (-3-sqrt(129))/4 + 3)`

`20/(-3-sqrt(129)) = 8/(9-sqrt(129)) + 2`

`-1.39296944... = -3.39296944... + 2color(white)(x)color(green)(sqrt())`