Question

Find the value for x and y if x+(xy)^1/2+y=14 and x^2+xy+y^2=84​

Answer 1

Answer:

$xy = 4$

Step-by-step explanation:

Taking the first equation , we have ,

$\implies x + (xy)^{\frac{1}{2}} + y = 14 \\\\\implies x + \sqrt{xy}+ y = 14 \\\\\implies x + y = 14 -\sqrt{xy}\\\\\implies (x+y)^2 = (14-\sqrt{xy})^2 \\\\\implies x^2+y^2+2xy = 196 + xy -28\sqrt{xy} \\\\\implies x^2+y^2+xy = 196 - 28\sqrt{xy}$

Now according to second equation,

$\\\\\implies 84 = x^2+xy+y^2$

Plug on this value ,

$\implies 84 = 196 -28\sqrt{xy}\\\\\implies -112 = -28\sqrt{xy}\\\\\implies \sqrt{xy}=\dfrac{112}{28}\\\\\implies \boxed{\boxed{ xy = 16 }}$

On substituting this value in terms of one variable of the equation you can get the values of x and y .