simultaneous equations Solve the simultaneous equations: x^2+y^2=200- xy x+y=20-(xy)^(1/2)
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The equations to be solved are:
x^2 + y^2 = 200 - xy ...(1)
x + y = 20 - (xy)^(1/2) ...(2)
(2)
=> (x + y)^2 = (20 - (xy)^(1/2))^2
=> x^2 + y^2 + 2xy = 400 + xy - 40(xy)^(1/2)
=> x^2 + y^2 + xy = 400 - 40(xy)^(1/2)
substiute in (1)
=> 200 = 400 - 40(xy)^(1/2)
=> 5 = 10 - (xy)^(1/2)
=> (xy)^(1/2) = 5
=> xy = 25
=> x = 25/y
substitute in x + y = 20 - (xy)^(1/2)
=> 25/y + y = 20 - (y*(25/y))^(1/2)
=> 25/y + y = 20 - (25)^(1/2)
=> 25/y + y = 20 - 5
=> 25/y + y = 15
=> y^2 - 15y + 25 = 0
y1 = 15/2 + sqrt (225 - 100) / 2
=> y1 = 15/2 + (sqrt 125)/2
y2 = 15/2 - (sqrt 125)/2
x1 = 25/y = 25/(15/2 + (sqrt 125)/2) = 15/2 - (sqrt 125)/2
x2 = 15/2 + (sqrt 125)/2
The solutions of the equations are: ( 15/2 + (sqrt 125)/2, 15/2 - (sqrt 125)/2) and (15/2 - (sqrt 125)/2, 15/2 + (sqrt 125)/2)
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Let a = x+y and b = sqrt(xy) => b^2 = xy
x^2 + y^2 = (x+y)^2 - 2xy
x^2 + y^2 = a^2 - 2b^2
We'll re-write the first equation, with respect to a and b:
x^2 + xy + y^2 = 200
a^2 - 2b^2 + b^2 = 200
We'll combine like terms and we'll get:
a^2 - b^2 = 200 (1)
We'll re-write the difference of squares:
a^2 - b^2 = (a-b)(a+b)
(a-b)(a+b) = 200 (2)
We'll change the 2nd original equation into:
x + sqrt(xy) + y = 20
a + b = 20 (3)
We'll substitute (3) in (2):
20(a-b) = 200
We'll divide by 20:
a - b = 10 (4)
We'll add (3) + (4):
a + b + a - b = 20 + 10
We'll combine and eliminate like terms:
2a = 30
a = 15
b = 20 - a
b = 20 - 15
b = 5
Now, we'll determine x and y:
x + y = 15
x*y = 25
We'll determine x and y knowing the sum and the product:
x^2 - 15x + 25 = 0
x = [15 + sqrt(225 - 100)]/2
The pairs (x,y) are: ((15 + 5sqrt5)/2;(15 - 5sqrt5)/2) ; ((15 - 5sqrt5)/2;(15 + 5sqrt5)/2)
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