how to convert one expression to another

Question

I equated them to each other, and they were equal, but .. how to make another from one

Answer 1

$$\frac{\sqrt3y+2x-2y}{\sqrt3y+2y} = \frac{\sqrt3y+2x-2y}{y\left(\sqrt3+2\right)}\left(\frac{\sqrt3-2}{\sqrt3-2}\right) = \frac{\left(\sqrt3y+2x-2y\right)\left(\sqrt3-2\right)}{-y}$$

Answer 2

The standard tricks to "deradicalize" are:

1) If you have $\sqrt a$ by itself in the denominator, such as $\frac M{\sqrt a}$, then multiply top and bottom by $\sqrt{a}$ to get $\frac {M\sqrt{a}}{\sqrt a\sqrt a}=\frac {M\sqrt{a}}{a}$

1b) If you have $m\sqrt a$ in the denominator, you can multiply top and bottom by $\sqrt a$ (the $m$ isn't relevant) to get $\frac {M}{m\sqrt a}=\frac {M\sqrt a}{ma}$.

2) If you have $\sqrt a$ as part of a sum (or difference) in the denominator, such as $\frac M{\sqrt a \pm k}$ then multiply by the complement[1] of the denominator, that is:

$\frac M{\sqrt a \pm k}=\frac {M(\sqrt a \mp k)}{(\sqrt a \pm k)(\sqrt a \mp k)} =\frac {M(\sqrt a \mp k)}{a - k^2}$

2a) If you have a $\sqrt a \pm \sqrt b$, two radicals, in a sum on the denominator, you can multiply by the complement in the same way.

$\frac M{\sqrt a \pm \sqrt b} = \frac {M(\sqrt a \mp \sqrt b)}{(\sqrt a \pm \sqrt b)(\sqrt a \mp \sqrt b)}=\frac {M(\sqrt a \mp \sqrt b)}{a-b}$.

(This assumes, of course that $a \ne b$. But if $a = b$ we'd have $\frac M{\sqrt a + \sqrt b} =\frac M{\sqrt a + \sqrt a} =\frac M{2\sqrt a}$ in which case you do 1b.)

2b) If you have $m\sqrt a \pm mk$ or $m\sqrt a \pm m\sqrt b$ in the denominator you can factor the $m$ out multiply top and bottom by the complement of $\sqrt a\pm k$ or $\sqrt a \pm \sqrt b$ which is $\sqrt a\mp k$ or $\sqrt a\mp \sqrt b$ and do as above.

2c) If you have $m\sqrt a \pm nk$ and $m,n$ have a common factor $d$ you can factor the $d$ out and ... do as above.

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So you have in the denominator $\sqrt 3y + 2y$. You can do 2b).

Factor out the $y$ to get $y(\sqrt 3 + 2)$ and mulitply top and bottom by the complement; $\sqrt 3 - 2$.

$\frac {\sqrt 3y + x-2y}{\sqrt 3y + 2y}=\frac {(\sqrt 3y + x-2y)(\sqrt 3- 2)}{y(\sqrt 3 + 2)(\sqrt 3- 2)}=\frac {(\sqrt 3- 2)(\sqrt 3y + x-2y)}{y(3 -4)}=\frac {(\sqrt 3- 2)(\sqrt 3y + x-2y)}{-y}=-\frac {(\sqrt 3- 2)(\sqrt 3y + x-2y)}{y}$

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[1] The complement of a sum $m \pm n$ is $m \mp n$. By multiplying together the sum and the difference: $(m+n)(m-n)= m\cdot m -m\cdot n + n\cdot m - n\cdot n = m^2 - n^2$ we get a "neat" difference between two squares and all the "middle stuff" cancels out.

If one, or both, of the terms is a radical that the radical is squared and ... isn't a radical any more:

$(\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = \sqrt a\sqrt a -\sqrt a\sqrt b +\sqrt a\sqrt b -\sqrt b\sqrt b = \sqrt a^2 -\sqrt b^2 = a-b$.