Aspire Faculty ID #10459 · Topic: NIMCET 2014 · Just now
NIMCET 2014

If x and y are positive real numbers satisfying the system of equations $x^{2}+y\sqrt{xy}=336$ and $y^{2}+x\sqrt{xy}=112$, then x + y is:

Solution

$x^2+y\sqrt{xy}=336 \quad ...(1)$ $y^2+x\sqrt{xy}=112 \quad ...(2)$ 
Dividing (1) by $\sqrt{x}$, 
$x\sqrt{x}+y\sqrt{y}=\frac{336}{\sqrt{x}}$ 

Dividing (2) by $\sqrt{y}$, 
 $y\sqrt{y}+x\sqrt{x}=\frac{112}{\sqrt{y}}$ 

 Since LHS are equal, $\frac{336}{\sqrt{x}}=\frac{112}{\sqrt{y}}$ 
 $\frac{\sqrt{x}}{\sqrt{y}}=3$ 
$\sqrt{x}=3\sqrt{y}$ 
 $\therefore x=9y$ Substituting in (2), 
 $y^2+9y\sqrt{9y^2}=112$ 
 $y^2+27y^2=112$ 
 $28y^2=112$ 
 $y^2=4$ 
 $\therefore y=2$ $x=9(2)=18$ 
 $\therefore \boxed{x+y=20}$
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