Bhaskara- $2008$

${ x }^{ 2 }+y\sqrt { xy } =336$

${ y }^{ 2 }+x\sqrt { xy } =112$

$\Rightarrow$ $\sqrt { x } (x\sqrt { x } +y\sqrt { y } )=336$ and $\sqrt { y } (y\sqrt { y } +x\sqrt { x } )=112$

$\Rightarrow$ $\frac { \sqrt { x } }{ \sqrt { y } } =\quad 336/112\quad =\quad 3\quad \Rightarrow \quad x=9y$

From ${ x }^{ 2 }+y\sqrt { xy } =336$ , we get

$81{ y }^{ 2 }+y\sqrt { 9{ y }^{ 2 } } =336$

$\quad \Rightarrow \quad 81{ y }^{ 2 }+3{ y }^{ 2 }=336$

$\quad \Rightarrow \quad 84{ y }^{ 2 }=336$

$\quad \Rightarrow \quad y=2$

Hence $x+y = 10y = 10*2 = 20$

My method was bit different and a bit lengthy..

Comparing equation 1 & 2 we can conclude that x>y

Replacing y by x in equation 1..

So value of x we get is 12.96

So, we can say that x>12.96

Similarly replacing x by y and we get y<7.48

As the final number is without decimals so √(xy) should be a perfect square..

Finding x & y such that xy are perfect square and satisfy any one equation..

We get 18 & 2