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Solve for x & y

$\sqrt x + y = 7$

$x + \sqrt y = 11$

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 $x = 9$

$y = {t^2} = 4$