# simultaneous equations Solve the simultaneous equations: x^2+y^2=200- xy x+y=20-(xy)^(1/2)

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### 2 Answers

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)**

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)