x+y+√z=148

Question

x+√y+z=82
√x+y+z=98

find x,y,z.

Please give full solution and how to find with complete logic.

Answer 1

z = (148-x-y)^2

x+√y + (148-x-y)^2 = 82
√x+y + (148-x-y)^2 = 98

now you have just two equations in x and y.

Or, assume integer solutions. Then x,y,z must all be perfect squares. Looking at the sums, it should be clear that y>x>z. Now just try a few small numbers.

Let z=9. Then you have

x+y=145
x+√y=73
√x+y=89

Now, with x and y both squares, 64+81=145 ...