105k views
4 votes
Find the value for x and y if x+(xy)^1/2+y=14 and x^2+xy+y^2=84​

User Ziko
by
6.6k points

1 Answer

4 votes

Answer:


xy = 4

Explanation:

Taking the first equation , we have ,


\implies x + (xy)^{(1)/(2)} + y = 14 \\\\\implies x + √(xy)+ y = 14 \\\\\implies x + y = 14 -√(xy)\\\\\implies (x+y)^2 = (14-√(xy))^2 \\\\\implies x^2+y^2+2xy = 196 + xy -28√(xy) \\\\\implies x^2+y^2+xy = 196 - 28√(xy)

Now according to second equation,


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

Plug on this value ,


\implies 84 = 196 -28√(xy)\\\\\implies -112 = -28√(xy)\\\\\implies √(xy)=(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 .

User Rboy
by
6.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.