$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ 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 :)
 A: 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 $x=13,y=10$
A: 
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$,$x+y=23$


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 $(x = 13, y = 10).$
A: Hint:
Let $\sqrt{x^2+12y}-\sqrt{y^2+12x}=1+2a$
$$\implies\sqrt{x^2+12y}=\dfrac{23+1+2a}2,\sqrt{y^2+12x}=11-a$$
$\implies11\ge a\ge-12\ \  \ \ (1)$
$$(12+a)^2-(11-a)^2=x^2-y^2-12(x-y)=(x-y)(x+y-12)$$
Replace the value of $x+y$ to find $x-y$
Consequently we find $x,y$ in terms of $a$
But we have $$x^2+12y=(12+a)^2$$
Replace the values of $x,y$ in terms of $a$ to from a quadratic equation $a$
Solve and remember $(1)$
A: $$\left(\sqrt{x^2+12y}+\sqrt{y^2+12x}\right)^2=(x+y+10)^2$$
$$\implies x^2+12y+y^2+12x+2\left(\sqrt{(x^2+12y)(y^2+12x)}\right)=x^2 + y^2 + 100+ 2xy+20x+20 y $$
$$\implies \sqrt{(x^2+12y)(y^2+12x)}=xy+4(x+y)+50 $$
$$\implies(x^2+12y)(y^2+12x)=(xy+142)^2 $$
$$\implies x^3 - \dfrac{35 x y}{3}+ y^3 - \dfrac{5041}{3}=0$$
$$ \implies (x+y)^3-3xy(x+y)-\dfrac{35 x y}{3}-\dfrac{5041}{3}=0$$
$$\implies 23^3-69xy-\dfrac{35 x y}{3}-\dfrac{5041}{3}=0$$
Finally, we have
$$\begin{cases} x+y=23 \\ xy=130 \end {cases}$$
$$\implies x=13, y=10 $$ and
$$\implies x=10, y=13.$$
A: WLOG
$$\sqrt{x^2+12y}=33\cos^2t\text{ and }\sqrt{y^2+12x}=33\sin^2t$$
$$(33\cos^2t)^2-(33\sin^2t)^2=x^2-y^2-12(x-y)=(x-y)(x+y-12)$$
$$x+y=23\implies x-y=99(\cos^2t-\sin^2t)=99(2\cos^2t-1)$$
$$\implies x=99\cos^2t-38\implies y=61-99\cos^2t$$
$$\implies(33\cos^2t)^2=x^2+12y=(99\cos^2t-38)^2+12(61-99\cos^2t)$$
Let $33\cos^2t=b$ $$\implies0=b^2-33b+272=(b-16)(b-17)$$
Hope you can take it from here?
A: The obvious (simple) solution is not repeated; instead we try to see the full picture.
Whenever there are radicals with polynomials of the form including $\pm$
$$ \pm \sqrt {P(x,y)}\pm \sqrt {Q(x,y)}= 1, $$
a higher degree polynomial should be recognized whose factors/ equations/constituents appear separately. They can  be graphed separately as below with different colors and labels of each part. $x\pm y=0$ lines and an asymptote are drawn included for reference:

EDIT1:
Continuous segments can be shown graphed together (it can be plotted in desmosas well) in Mathematica.
ContourPlot[{(Sqrt[x^2 + 12 y] +Sqrt[y^2 + 12 x] +33) *
    (Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] +33)*
    (-Sqrt[x^2 + 12 y] +Sqrt[y^2 + 12 x] +33)*
    (+Sqrt[x^2 + 12 y] +Sqrt[
       y^2 + 12 x] -33)*(+Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] -33)
    (+Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] -33) == 0
  }, {x, -p, p}, {y, -p, p}, GridLines -> Automatic, 
 ContourStyle -> {Thick, Red}]

Six factors and a product polynomial plot with uncut segments are shown.

A neat rational cubic product polynomial equation is cumbersome and unless seen in that form I feel as having not taken to the finish. A symmetric graph with respect to y-axis could be obtained by $45^{\circ}$ CW rotation.
