Question

so, how many can solve
sqrt{x} + y = 11
x + sqrt{y} = 7
sole this by process and steps..

Answer 1

Just looking at this gives a logical answer of x=4, y=9

Answer 2

Without solving any equations, we can find this solution.

x and y have to be square integers and x has to be less than 11 and y less than 7. The only square integers less than 11 is 9, 4, 1 and 0 and the only squares less than 7 are 4,1 and 0.

In the 1st equation, if x was 9 then y should be 8, which isn't a perfect square. If x was 4 then y is 9, which IS a perfect square.

To check, look at the 2nd equation, 4 + 3 is 7, so that must be the solution.

Answer 3

any1 can guess d answr... I askd 2 solv it by step..

Answer 4

$$
\sqrt{x} + y = 11\implies\sqrt{x}=11-y\implies x=(11-y)^2=121-22y+y^2 \\
x + \sqrt{y} = 7\implies\sqrt{y}=7-x\implies y=(7-x)^2= 49-14x+x^2\\
$$
Add \(x+y\):
$$
x+y=121-22y+y^2 + 49-14x+x^2\\
0=121-22y-y+y^2 + 49-14x-x+x^2\\
0=170-15x-23y+x^2+y^2\\
$$
Which is a circle of radius, \(\sqrt{\cfrac{37}{2}}\):
$$
(x-\cfrac{15}{2})^2+(y-\cfrac{23}{2})^2=\cfrac{37}{2}
$$
http://en.wikipedia.org/wiki/Circle#Equations

Some integer solutions include:
4,9
4,14
5,8
5,15
10,8
10,15
11,9
11,14

Hope this helps.

Answer 5

@ybarrap: Your answer is wrong. For example pair x=4, y=14 doesn't solve the original equations.

Only pair x=4, y=9 solves the equations.