<s>[INST] $\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$ 
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$

The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.


The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)
 [/INST] Don't rush to square equations to get rid of square roots.  It's almost always more work and you end up missing opportunities for simplification.  Instead eliminate square roots by substitution.  Here, let $a\ge0$ and $b\ge0$ be the values of the two square roots:
$$
\begin{cases}
x+y=23\\
a+b=33\\
a^2 = x^2 + 12y\\
b^2 = y^2 + 12x
\end{cases}
$$
While this turns your two equations into four equations, it makes simplifications easier to spot.  Subtracting the last two equations will let you factor $a^2 - b^2$ and $x^2 - y^2$ and make use of the fact that $x+y$ and $a+b$ are known:
$$33(a-b) = 23(x-y) - 12(x-y)$$
$$3(a-b) = x-y$$
Now this becomes very simple.  Let $z = a-b$ and solve for $z$.   For example, express $x$, $y$, $a$ in terms of $z$ and substitute into $(a-x)(a+x) = 12y$.  Everything will neatly cancel out and you will get $z^2 = 1$ or $a=b\pm1$ and $x=y\pm3$.  The two solutions are $x=10, y=13$ and $x=13,y=10$
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