Equation in 2 variables

I gave up doing normal stuffs we normally do, so my last effort is the brute force with little help from calculators. :<

From $\sqrt{y} + x = 11$ (2)

$\sqrt{y} = 11 - x$

Substitute into (1)

$x^{2} - 22x + \sqrt{x} + 121 = 7$

Let $a = \sqrt{x}$

$a^{4} - 22a^{2} + a + 114 = 0$

$(a-3)(a^{3}+3a^{2}-13a-38)= 0$

This right bracket degree 3 equation gives you 2 negative roots and 1 positive roots, which doesn't work on both equations for x and y. I tried by calculator, sorry. :(

So $a = 3$

$x = 9$

$y = 4$

Therefore, $x+y = \boxed{13}$

Subtracting the two equations we get

$y-x+\sqrt{x}-\sqrt{y}=\,-4$

$\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-1\right)=4$ .

Now since $x,y\;$ $\in{\mathbb{N}}$ , so

$\sqrt{x}+\sqrt{y}>1\rightarrow\;\sqrt{x}\geq\sqrt{y}\;\;$ $($ Since their product is positive $)$ .

So we need to check for two cases viz. $\left(\sqrt{x}-\sqrt{y},\sqrt{x}+\sqrt{y}-1\right)=\left(1,4\right),\left(2,2\right)$ .

We can see that the latter doesn't yield integral solutions. So solving the former gives $\left(x,y\right)=\left(9,4\right)$ .

So $x+y=\boxed{13}$ .

$\sqrt{x} + y = 7$

=> y - 7 = $\sqrt{x}$

=> $(y - 7)^2$ = x

=> $y^2$ - 14y + 49 = x

Now, $\sqrt{y}$ + x = 11

=> $\sqrt{y}$ + $(y - 7)^2$ = 11

=> $y^2$ - 14y + $\sqrt{y}$ + 38 = 0

Taking $\sqrt{y}$ = t

$t^4$ - 14 $t^2$ + t + 38 = 0

Applying remainder theorem

t = 2 will be one of the roots

So, (t - 2)( $t^3$ - 2 $t^2$ - 10t - 19) = 0

Again applying remainder theorem on the second part we can easily see that no integer will be the root as the constant term is a prime no. so it's factors will be only 1 and 19 or -1 and -19 which do not satisfy the 2nd part

This way only t = 2 is the integer solution so y = 4

and x = 9

Let $a=\sqrt{x}$ and $b=\sqrt{y}$ . The system then becomes: $a+b^2=7\\ a^2+b=11$

Subtract the equations: $11-7=(a^2-b^2)-(a-b) \implies 4=(a-b)(a+b-1)$ , which means that we either have $\textcolor{#D61F06}{(a-b)}\cdot(a+b-1)=\textcolor{#D61F06}{2}\cdot2$ , which is not a solution to the system, or $\textcolor{#D61F06}{(a-b)}\cdot(a+b-1)=\textcolor{#D61F06}{1}\cdot4$ , which is. This last case yields $a=3$ and $b=2$ , that is, $x=a^2=9$ and $y=b^2=4$ , so $x+y=\boxed{13}$ .

Let\quad \sqrt { y } =t.\ So,\quad we\quad get\quad an\quad equation,\ { t }^{ 4 }-14.{ t }^{ 2 }+t+38=0.\ By\quad solving\quad the\quad quartic\quad equation\quad by\quad depressing\quad it,\ we\quad get\quad 2\quad positive\quad solutions\quad for\quad t.\ either\quad t=2\quad or\quad t=3.1313125\ But,\quad 7-{ t }^{ 2 }>0\ so,\quad t\quad cannot\quad be\quad more\quad than\quad 3,\ so\quad t=2,\quad y=4,\quad x=9\ Result:\quad x+y=13