Dare you to use substitution 2

From the first equation $\sqrt{2x}+y = 28$ , we note that for $\sqrt{2x}$ to be an integer, $\sqrt{2x} = \sqrt{2\times 2n^2}$ ) and $x = 2n^2$ , where $n = 1, 2, 3...,13$ .

Substitute into the first equation:

$\sqrt{4n^2} + y = 28\quad \Rightarrow y = 28 - 2n$ , this means that $y$ is even and $y \le 26$ .

From the second equation $x^2 - y^3 - \sqrt{y}^5 = 64$ , we note that $\sqrt{y}$ have to be an integer and since $y$ is even, $\sqrt{y} = \sqrt{4m^2}$ and $y = 4m^2$ , where $m = 1, 2,...$ . Since $y \le 26$ , $y = 4$ or $16$ .

From the second equation, when $y = 4$ ,

$x^2 - 4^3 - \sqrt{4}^5 = 64$

$\Rightarrow x^2 = 64 + 64 + 32 = 160 \quad \Rightarrow x = 4\sqrt{10}$ which is not an integer.

When $y = 16$ ,

$\Rightarrow x^2 = 64 + 16^3 + \sqrt{16}^5 = 64 + 4096 + 1024 = 5184$

$\Rightarrow x = \sqrt{5184}=72$

$\therefore x+y= 72+16= \boxed{88}$

$(i)Given\quad x\& y\in { I }^{ + }\\ \\ \\ (ii)By\quad inspection\quad of\quad both\quad { eq }^{ n },\\ \\ 2x\quad \& \quad y\quad are\quad whole\quad square\quad numbers\\ \\ \\ (iii)By\quad { 2 }^{ nd\quad }{ eq }^{ n }\quad \\ \\ { x }^{ 2 }=64+{ y }^{ 3 }+{ y }^{ 5/2 }\\ \\ \Rightarrow { x }^{ 2 }\ge 64\quad (as\quad y\quad is\quad +ve)\\ \\ \Rightarrow x\ge 8\quad (as\quad x\quad is\quad +ve)\\ \\ \\ by\quad { 1 }^{ st }\quad { eq }^{ n },\quad \\ \\ y=28-\sqrt { 2x } \\ \\ \Rightarrow y\le 24\quad (as\quad x\ge 8)\\ \\ so\quad possible\quad values\quad of\quad y\quad are\quad 1,4,9,\& 16.\\ \\ \\ Use\quad hit\quad \& \quad trial\Rightarrow only\quad one\quad { sol }^{ n }\quad (72,16)\\ \\ x+y=72+16=88\\$