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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x^2 + xy + y62 = 200...............(1)

x + (xy)^1/2 + y = 20...............(2)

First we will re-wrtie equation (2).

We will move (xy)^1/2 to the right sides.

==> (x+ y) = 20 - (xy)^1/2

Now we will square both sides.

==> (x+ y)^2 = (20- (xy)^1/2) ^2

==> x^2 = 2xy + y62 = 400 - 40(xy)^1/2) +xy

Now subtract xy from both sides.

==> x^2 + xy + y^2 = 400 - 40(xy)^1/2)

But, from equation (1) , we know that x^2 + xy + y^2 = 200.

==> 200 = 400 - 40(xy)^1/2)

==> -200 = -40(xy)^1/2

==> (xy)^1/2 = 200/40 = 5

==> xy = 25.............(3)

Now we will substitute in (1) and (2).

==> x^2 + y^2 + 25 = 200

==> x^2 = y^2 = 175 ............(4)

==> x+ 5 + y 20

==> x+ y = 15

==> y= 15-x ...............(5)

Now substitute on (3)

==> x*y = 25

==> x( 15-x) = 25

==> 15x -x^2 = 25

==> x^2 - 15x + 25 = 0

==> x1=( 15+ 5sqrt5)/2  ==> y1= ( 15-5sqrt5)/2

==> x2= (15-5sqrt5) /2 ==> y2= (15+5sqrt5)/2

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

x^2 + xy + y^2 = 200

and x + (xy) ^ (1/2) + y = 20

Now x^2 + xy + y^2 = 200

=> (x + y) ^2 - 2xy + xy = 200

=> (x + y) ^2 - xy = 200

Also, x + (xy) ^ (1/2) + y = 20

=> x + y + sqrt xy = 20

Let x + y= X and xy = Y

So we have: X^2 - Y = 200

and X + sqrt Y = 20

(X + sqrt Y)^2 = 20^2

=> X^2 + Y + 2X sqrt Y = 400

Now sqrt Y = 20 - X and Y = X^2 - 200

substituting these in X^2 + Y + 2X sqrt Y = 400

=> X^2 + X^2 - 200 + 2X (20 - X) = 400

=> 2X^2 - 200 + 40 X - 2X^2 = 400

=> - 200 + 40 X = 400

=> 40 X = 600

=> X = 600 / 40 = 15

Y = X^2 - 200

=> Y = 225 - 200 = 25

So x+y = 15 and xy = 25

x+y = 15 => x = 15 - y

substitute this in xy = 25

=> (15 - y) y = 25

=> 15y - y^2 - 25 = 0

=> y^2 - 15y + 25 = 0

y1 = [15 + sqrt (225 - 100)]/2

= (15 + 5sqrt5)/2

y2 = (15 - 5sqrt5)/2

x1 = 15 - (15 + 5sqrt5)/2 = (15 - 5sqrt5)/2

x2 = (15 + 5sqrt5)/2

Therefore the values x and y can take are ((15 - 5sqrt5)/2, (15 + 5sqrt5)/2) and ((15 + 5sqrt5)/2, (15 - 5sqrt5)/2)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll note 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

x = (15 + 5sqrt5)/2 and y = (15 - 5sqrt5)/2

or

x = (15 - 5sqrt5)/2 and y = (15 + 5sqrt5)/2

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