Putting $y = {(11 - x)^2}$ obtained from the 2nd equation into the 1st equation,
We have, $\sqrt x + {(11 - x)^2} = 7$
$ \Rightarrow x = {[7 - {(11 - x)^2}]^2}$
$ = 49 + {(11 - x)^4} - 14{(11 - x)^2}$
Let, $11 - x = t$
So, $11 - t = 49 + {t^4} - 14{t^2}$
$ \Rightarrow {t^4} - 14{t^2} + t + 38 = 0$
t=2 satisfies the above equation. So, using factor theorem
${t^3}(t - 2) + 2{t^2}(t - 2) - 10t(t - 2) - 19(t - 2) = 0$
$ \Rightarrow (t - 2)({t^3} + 2{t^2} - 10t - 19) = 0$ ........... (A)
Let, $f(t) = {t^3} + 2{t^2} - 10t - 19$
$t = 11 - x = \sqrt y \ge 0$
$y = 7 - \sqrt x , \Rightarrow y \le 7$
So, $t = \sqrt y \le \sqrt 7 $
Thus, $0 \le t \le \sqrt 7 $
Or, $0 \le t < 3$
Let us investigate the behaviour of f in the interval [0, 3].
f(0) = -19
f(1) = -26
f(2) = -23
f(3) = -4
The function decreases initially, then increases but is unable to cut the t-axis in the interval [0, 3).
t = 2 is the only solution as obtained earlier as per (A).
Now, $11 - x = 2$ or $x = 9$
$y = {t^2} = 4$