# How to show that algebraic fractions can be simplified?

## Show that $\frac{5}{x - 3} + \frac{2}{x + 1} = 3$ can be simplified to $3 {x}^{2} - 13 x - 8 = 0$

May 15, 2016

The idea is that you can multiply both sides of the equation by the lowest common multiple of the denominators. This might sound confusing but you'll see why we do this.

So the denominators are $x - 3$, $x + 1$ and $1$ (consider that you can express $3$ as $\frac{3}{1}$.

If we multiply both sides by $\left(x - 3\right) \left(x + 1\right)$ (obviously we don't need to multiply by the $1$ as it will have no effect).

$\frac{5}{x - 3} + \frac{2}{x + 1} = 3$

Will become:

$\left(x - 3\right) \left(x + 1\right) \frac{5}{x - 3} + \left(x - 3\right) \left(x + 1\right) \frac{2}{x + 1} = \left(x - 3\right) \left(x + 1\right) 3$
(You may notice that things are badly ordered here but I'm just trying to show what I've done).

By cancelling factors/reordering where necessary:

$5 \left(x + 1\right) + 2 \left(x - 3\right) = 3 \left(x - 3\right) \left(x + 1\right)$

$5 x + 5 + 2 x - 6 = 3 \left({x}^{2} + x - 3 x - 3\right)$

$7 x - 1 = 3 {x}^{2} - 6 x - 9$

$3 {x}^{2} - 13 x - 8 = 0$, as required.

If you wanted to go any further you could solve this using the quadratic formula for solutions of x.
For further development, you could use a graphing app to show that what you've found is the points of intersection of the two lines:

$y = \frac{3}{x - 3} + \frac{2}{x + 1}$ and $y = 3$

A further point on the above. It isn't necessary to multiply through by the LCM of the denominators every time. If you just multiply through by the denominators and then work through from there, it will always remove fractions and make it easier to solve.

If we had $2 x + 2$ and $x + 1$ as denominators then it would suffice to multiply through by $2 x + 2$ since this is the lowest common multiple of the denominators ($2 \left(x + 1\right) = 2 x + 2$).