<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] You finished with $$\log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3$$ Go to natural logarithms (the only ones I know !); this will then write $$\frac{\log (4-2 x)}{\log (x+1)}+\frac{\log (5-x)}{\log (x+1)}=3$$ Multiply by the denominator appearing in the lhs (hoping that $x\neq -1$) and simplify. 
So, $$\log (4-2 x)+\log (5-x)=3\log(x+1)$$ that is to say $$(4-2x)\times (5-x)=(x+1)^3$$ Expand and simplify to get $$x^3+x^2+17 x-19=0$$ where $x=1$ is an obvious solution. Perform the long division to get $$x^3+x^2+17 x-19=(x-1)(x^2+2 x+19)=0$$ The quadratic term does not show real solutions (in the complex domain, they would be $(x_{1,2}=-1\pm3 i \sqrt{2})$.
So $x=1$ is your solution and you were about to get it.
Happy New Year !!!!
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