SOLUTION: solve the following system of equations: x + y + square root(xy) = 28 x^2 + y^2 + xy = 336

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Question 348952: solve the following system of equations:
x + y + square root(xy) = 28
x^2 + y^2 + xy = 336

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
solve the following system of equations:
x + y + sqrt%28xy%29 = 28
x^2 + y^2 + xy = 336
:
Simplify the 1st equation to use for elimination
x + y = 28+-+sqrt%28xy%29
square both sides
x%5E2+%2B+2xy+%2B+y%5E2 = 784+-+56sqrt%28xy%29+%2B+xy
subtract xy from both sides
x%5E2+%2B+2xy+-+xy%2B+y%5E2 = 784+-+56sqrt%28xy%29
x%5E2+%2B+xy%2B+y%5E2 = 784+-+56sqrt%28xy%29
:
Use for elimination with 2nd equation
x%5E2+%2B+y%5E2%2B+xy = 784+-+56sqrt%28xy%29
x%5E2+%2B+y%5E2+%2B+xy+=+336
---------------------------Subtraction eliminates everything on the left, we have:
0+=+448+-+56sqrt%28xy%29
56sqrt%28xy%29+=+448
divide both sides by 56, results:
sqrt%28xy%29+=+8
square both sides
xy = 64
y = 64%2Fx
:
Substitute 64%2Fx for y in the 1st equation
x + y + sqrt%28xy%29 = 28
x + 64%2Fx + sqrt%28x%2864%2Fx%29%29 = 28
x's inside the radical cancel
x + 64%2Fx + sqrt%2864%29%29 = 28
x + 64%2Fx + 8 = 28
x + 64%2Fx = 28 - 8
x + 64%2Fx = 20
multiply by x
x^2 + 64 = 20x
A quadratic equation
x^2 - 20x + 64 = 0
Factors to
(x-16)(x-4)
x = 16
x = 4
:
Find y:
When x=16: y = 64/16 = 4
When x=4: y = 64/4 = 16
:
:
Check solution in the 1st equation
16 + 4 + sqrt%2816%2A4%29 = 28
20 + sqrt%2864%29 = 28; confirms our solutions; x=16, y-4 and x=4, y=16