For integers x and y, we know sqrt(x)+ sqrt(y) = 16 and x-y =-128. what is the value of x. 

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We have two equations

`sqrt(x)+sqrt(y) = 16`  --------->1

`x -y = -128` --------->2

we can rewrite 2 as,

`(sqrt(x)+sqrt(y))(sqrt(x)-sqrt(y)) = -128`

16(sqrt(x)-sqrt(y)) =-128

`sqrt(x)-sqrt(y) = -8` we shall call this equation as 3.

Now if we add 1 to 3 you get,

`2sqrt(x) = 16 -8 = 8`

sqrt(x) =...

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We have two equations

`sqrt(x)+sqrt(y) = 16`  --------->1

`x -y = -128` --------->2

we can rewrite 2 as,

`(sqrt(x)+sqrt(y))(sqrt(x)-sqrt(y)) = -128`

16(sqrt(x)-sqrt(y)) =-128

`sqrt(x)-sqrt(y) = -8` we shall call this equation as 3.

Now if we add 1 to 3 you get,

`2sqrt(x) = 16 -8 = 8`

sqrt(x) = 4

x = 16.

then substituting 2, you get,

16-y=-128

y = 16+128 = 144.

 

Therfore answer is x= 16 and y = 144.

 

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