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$X = Y + Z$ Where $Y$ and $Z$ are independent Gaussians with zero means and known variances.

How do you simplify the left side of this to equal the right side?

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I think I see why $Cov(Y+Z,Y+Z)$ equals $\sigma_{Y}^2+\sigma_{Z}^2$, but don't know why its in the denominator or how the numerator is obtained.

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It is a mistake somewhere in the book or wherever you got the problem. It should be:

$$\frac{\text{Cov}(Y,Y+Z)}{\text{Cov}(Y+Z,Y+Z)}x=\ldots$$ since: $$\require{cancel}\text{Cov}(Y,Y+Z)=\text{Cov}(Y,Y)+\cancel{\text{Cov}(Y,Z)}=\sigma_Y^2$$ and the denominator is as you said: $$\text{Cov}(Y+Z,Y+Z)=\text{Cov}(Y,Y)+\cancel{\text{Cov}(Y,Z)}+\cancel{\text{Cov}(Z,Y)}+\text{Cov}(Z,Z)=\sigma_Y^2+\sigma_Z^2$$

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    Thanks this must have been an error in the solution. Does the fact that Y and Z are independent imply that $Cov(Y,Z)$ is zero? – 2017-02-11
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    Yes, independent implies uncorrelated (if the second moment exists). – 2017-02-11
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    Ah ok I always mix that up – 2017-02-11
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    hey sorry I thought I understood why $\text{Cov}(Y,Y+Z)=\text{Cov}(Y,Y)+\text{Cov}(Y,Z)$, but now I can't remember. Would you mind breaking this part down too when you get a chance? – 2017-02-11