# How do you solve \frac { x + 2} { x } = \frac ( x - 1} { x } - \frac { 4x + 2} { x ^ { 2} - 3x }?

Jan 25, 2018

$x = 1$

(the solution $x = 0$ is extraneous)

#### Explanation:

First, move the $\frac{x - 1}{x}$ to the left side.

$\frac{x + 2}{x} = \frac{x - 1}{x} - \frac{4 x + 2}{{x}^{2} - 3 x}$

$\frac{x + 2}{x} - \frac{x - 1}{x} = - \frac{4 x + 2}{{x}^{2} - 3 x}$

Now you can combine the numerators on the left side since the denominators are the same.

$\frac{\left(x + 2\right) - \left(x - 1\right)}{x} = \frac{- \left(4 x + 2\right)}{{x}^{2} - 3 x}$

$\frac{3}{x} = \frac{- \left(4 x + 2\right)}{{x}^{2} - 3 x}$

Now we can just cross multiply and solve it like a quadratic:

$3 \left({x}^{2} - 3 x\right) = - x \left(4 x + 2\right)$

$3 {x}^{2} - 9 x = - 4 {x}^{2} - 2 x$

$7 {x}^{2} - 7 x = 0$

${x}^{2} - x = 0$

$x \left(x - 1\right) = 0$

$\therefore \text{ " x = 0 " " or " } x = 1$

So these should be our two solutions. But there's one last step - we need to plug these solutions back into the original equation to make sure they aren't extraneous (i.e. they produce an indeterminate form like $\frac{0}{0}$ instead of a real number).

Here's the original equation:

$\frac{x + 2}{x} = \frac{x - 1}{x} - \frac{4 x + 2}{{x}^{2} - 3 x}$

Let's check $x = \textcolor{red}{0}$

(color(red)0+2)/color(red)0 stackrel(?color(white)"=")(=) (color(red)0 - 1)/color(red)0 - (4(color(red)0)+2)/(color(red)0^2-3(color(red)0))

This already has a lot of dividing by zero, so unfortunately it is NOT a real solution since it will produce undefined values.

Now, let's check $x = \textcolor{b l u e}{1}$

(color(blue)1+2)/color(blue)1 stackrel(?color(white)"=")(=) (color(blue)1 - 1)/color(blue)1 - (4(color(blue)1) + 2)/(color(blue)1^2-3(color(blue)1))

3/1 stackrel(?color(white)"=")(=) 0/1 - 6/(-2)

3 stackrel(?color(white)"=")(=) -(-3)

3 stackrel(color(limegreen)sqrt"" color(white)(=))(=) 3

So our only solution is $x = 1$