Transformation using matrices (algebra help)

**RAz** · May 8th 2009, 10:25 PM

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)

**Gamma** · May 8th 2009, 11:29 PM

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:

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

**RAz** · May 9th 2009, 12:35 AM

Thankyou so much!

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