$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ 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 :)


$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$,$x+y=23$


The obvious way to solve it is by substituting for one variable.

No, cart before the horse.  The first step is to move one of the radicals to the RHS and square both sides.  This gives:
$$(x^2 + 12y) = 1089 + (y^2 + 12x) - 66\sqrt{y^2 + 12x}.\tag1$$
At this point, there are two key factors:

*

*When you square both sides, you will generate an equation such that some of its solutions may not satisfy the original equation.

*You are looking for some elegant way of utilizing that $(x + y) = 23.$
I experimented with various tries and then decided on the following:

*

*$(x^2 + 12y) = x^2 + 12[23 - x] = x^2 - 12x + 276.$


*$(y^2 + 12x) = [23-x]^2 + 12x = 529 - 46x + x^2 + 12x = x^2 - 34x + 529.$
Therefore, equation (1) above simplifies to
$$(x^2 - 12x + 276) = 1089 + (x^2 - 34x + 529) - 66\sqrt{x^2 - 34x + 529} \implies $$
$$(1342 - 22x) = 66\sqrt{x^2 - 34x + 529} \implies $$
$$(61 - x) = 3\sqrt{x^2 - 34x + 529} \implies $$
$$x^2 - 122x + 3721 = 9[x^2 - 34x + 529] \implies $$
$$(8x^2 - 184x + 1040) = 0 \implies$$
$$x = \left(\frac{1}{16}\right) \times
\left[184 \pm \sqrt{33856 - \left(4 \times 8 \times 1040\right)}\right]$$
$$= \left(\frac{1}{16}\right) \times
\left[184 \pm \sqrt{576}\right]$$
$$= \left(\frac{1}{16}\right) \times
\left[184 \pm 24\right] \implies x \in \{10, 13\}.$$
Here, you have to check each candidate value of $x$ against the original equation.  Doing so, you realize that there are in fact two solutions: $(x=10, y=13)$ and $(x = 13, y = 10).$
