Kinetic energy of daughter nucleus and alpha particle from alpha decay

[Doscience]

Homework Statement

In \alpha decay a nucleus X at rest decays to a daughter nucleus Y and an \alpha particle. Conservation of momentum and kinetic energy gives:
M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0

\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q

Where the Q value is the available energy found through Q=(M(X)-M(Y)-M(\alpha))c^{2}

Show the kinetic energy of the two decay products are given by

E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q

E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q

Homework Equations

\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q

M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0

Q=(M(X)-M(Y)-M(\alpha))c^{2}

E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q

E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q

The Attempt at a Solution

I have tried rearranging the energy equation to get E^{\alpha}_{k}=Q-\frac{1}{2}M_{Y}v_{Y}^{2}
since
E^{\alpha}_{k}=\frac{1}{2}M_{\alpha}v_{\alpha}^{2}

Then rearranging the conservation of momentum equation and substituting in for various variables but I can't get anything that looks like the required expressions. I know this is a fairly simple algebra exercise but I just can't figure out what to do so any advice or suggestions would be appreciated.

 

[Doscience]

I managed to figure this out. I rearranged the expression for Q to get E^{\alpha}_{K}like in my initial attempt but then I took Q out as a common factor which was the key. It was then just a matter of doing about 3 pages of algebra on what was inside the brackets.