<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] $$\left(\sqrt{x^2+12y}+\sqrt{y^2+12x}\right)^2=(x+y+10)^2$$
$$\implies x^2+12y+y^2+12x+2\left(\sqrt{(x^2+12y)(y^2+12x)}\right)=x^2 + y^2 + 100+ 2xy+20x+20 y $$
$$\implies \sqrt{(x^2+12y)(y^2+12x)}=xy+4(x+y)+50 $$
$$\implies(x^2+12y)(y^2+12x)=(xy+142)^2 $$
$$\implies x^3 - \dfrac{35 x y}{3}+ y^3 - \dfrac{5041}{3}=0$$
$$ \implies (x+y)^3-3xy(x+y)-\dfrac{35 x y}{3}-\dfrac{5041}{3}=0$$
$$\implies 23^3-69xy-\dfrac{35 x y}{3}-\dfrac{5041}{3}=0$$
Finally, we have
$$\begin{cases} x+y=23 \\ xy=130 \end {cases}$$
$$\implies x=13, y=10 $$ and
$$\implies x=10, y=13.$$
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