# Thread: Transformation using matrices (algebra help)

1. ## Transformation using matrices (algebra help)

I'm stuck on simplifying this:

$\frac{-1}{2}x' + \frac{\sqrt3}{2}y' = \frac{-\sqrt3}{2}x' - \frac{1}{2}y' + 4$

The negative is apart of the entire fraction, not just the numerator.

I think I have to apply the distributive law and rationalise the denominator, but I've never been any good at fractions to be honest.

(This is for matrix rotation, if you are curious)

2. Originally Posted by RAz
I'm stuck on simplifying this:

$\frac{-1}{2}x' + \frac{\sqrt3}{2}y' = \frac{-\sqrt3}{2}x' - \frac{1}{2}y' + 4$

The negative is apart of the entire fraction, not just the numerator.

I think I have to apply the distributive law and rationalise the denominator, but I've never been any good at fractions to be honest.

(This is for matrix rotation, if you are curious)
$\frac{-1}{2}x' + \frac{\sqrt3}{2}y' = \frac{-\sqrt3}{2}x' - \frac{1}{2}y' + 4$

$\frac{-1}{2}x' + \frac{\sqrt3}{2}x' = - \frac{1}{2}y' -\frac{\sqrt3}{2}y' + 4$

$\frac{\sqrt3-1}{2}x' = -\frac{\sqrt3+1}{2}y' + 4$

$\frac{2}{\sqrt3-1}\frac{\sqrt3-1}{2}x' = \frac{2}{\sqrt3-1}\left(-\frac{\sqrt3+1}{2}y' + 4 \right)$

$x' = -\frac{\sqrt3+1}{\sqrt3 - 1}y' + \frac{8}{\sqrt3 - 1}$

rationalize denominator:

$x' = -\frac{4+2\sqrt3}{4}y' + \frac{8\sqrt3+8}{4} = -(1 + \frac{\sqrt3}{2})y' + 2(\sqrt3 +1)$

3. Thankyou so much!