<s>[INST] Solve these equations simultaneously 
Solve these equations simultaneously:
  $$\eqalign{
  & {8^y} = {4^{2x + 3}}  \cr 
  & {\log _2}y = {\log _2}x + 4 \cr} $$

I simplified them first:
$\eqalign{
  & {2^{3y}} = {2^{2\left( {2x + 3} \right)}}  \cr 
  & {\log _2}y = {\log _2}x + {\log _2}{2^4} \cr} $
I then had:
$\eqalign{
  & 3y = 4x + 6  \cr 
  & y = x + 16 \cr} $
Solving:
$\eqalign{
  & 3\left( {x + 16} \right) = 4x + 6  \cr 
  & 3x + 48 = 4x + 6  \cr 
  & x = 42  \cr 
  & y = \left( {42} \right) + 16  \cr 
  & y = 58 \cr} $

This is the wrong answer, I would like to understand where I went wrong so I dont make the same mistake again, your help is greatly appreciated, thanks!
 [/INST] The equation wil be $\log_2 y= \log_2 x + \log_2 2^4$.
so $y$ will then be equal to $16x$.
So $4x$ will be $\dfrac y4$.
Substituting,so $3y=\dfrac y4+6$
$\dfrac {11y}4= 6$
So $y=\dfrac {24}{11}$. And $x=\dfrac {3}{22}$.
</s>