<s>[INST] Simultaneous Equations (Stuck on the algebra) 
Question: Solve the following simultaneous equations for real values of x and y
  $$
 \left\{
  \begin{array}{l}
   9^{2x+y} - 9^x \times 3^y = 6 \\
   \log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
  \end{array}
 \right.
$$

What I have attempted; for the first equation
$$
 9^{2x+y} - 9^x \times 3^y = 6 \\
 9^{2x+y} - 3^{2x} \times 3^y = 6 \\
 9^{2x+y} - 3^{2x+y} - 6 = 0
$$
Let $z = 3^{2x+y}$. Then
$$
 z^2 - z - 6 = 0
 \iff (z-3)(z+2) = 0
 \iff z = 3, z \ne -2
$$
where
$$
 3^{2x+y} = 3 \iff y = 1-2x.
$$
Now using the second equation I get
$$
 \log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
$$
Substituting $y = 1-2x$
$$
 \log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3.
$$
Now this is the part I am stuck on , how do I solve for $x$ algebraically? 
 [/INST] Writing the last  equation 
$$ \log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3 $$
as
$$ \log_{x+1}[(4-2x)(5-x)] = 3 $$
you can get both term as exponents of $x+1$, so that
$$ (4-2x)(5-x) = (x+1)^3 $$
This equation can be easily solved, leading to $x=1$.
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