How do i solve each system of nonlinear equations?

Question

2xy + y^2 = 16
3y^2 - xy = 2

Answer 1

I would make s=xy and work it out...

Multiply by 2 the second eq to eliminate "s"

2s + y^2 = 16
-2s+6y^2=4
__________
7y^2 = 20
y^2= 20/7
y= +- 2 sqroot (5/7)

then working with second eq
s= 3y^2 - 2
s= 60/7 -2
s= 46/7

then
xy = 46/7

and working with both values of y
y= 2 sqrt (5/7) = 2/7 sqrt(35).... this to simplificate rationalizing
x(2/7 sqrt(35))= 46/7
x= 23 / sqrt 35

and with y= -2/7 sqrt(35)

xy= 46/7
x(-2/7 sqrt(35))=46/7
x= - 23 / sqrt 35

so... result are
x,y
23 / sqrt 35 ; 2/7 sqrt(35) and
-23 / sqrt 35 ; -2/7 sqrt(35)

Answer 2

2xy + y^2 = 16
3y^2 - xy = 2

One thing you can do is solve for x in the first equation, and then plug it into the second. Let's manipulate the first equation.

2xy + y^2 = 16
2xy = 16 - y^2
x = (16 - y^2)/[2y]

Which we now plug into the second equation.

3y^2 - xy = 2
3y^2 - {(16 - y^2)/[2y]}y = 2

Notice the y cancels out the y in the denominator. This will give us:

3y^2 - {(16 - y^2)/2} = 2. Multiplying both sides by 2,
3y^2 - (16 - y^2) = 4. Distributing the minus,
3y^2 - 16 + y^2 = 4. Combining like terms,
4y^2 - 16 = 4. Bringing the -16 over to the right hand side,
4y^2 = 20. Dividing both sides by 4,
y^2 = 5. Taking the square root of both sides yields plus or minus sqrt(5).

y = +/- sqrt(5)

Now that we have *two* answers for y. We have to plug BOTH values in and get values for x. We already solved for x in the first equation (it was x = (16 - y^2)/(2y) ) so all we have to do is plug in each value individually.

y = sqrt(5):
x = (16 - [sqrt(5)]^2)/(2sqrt(5))) = [16 - 5]/(2sqrt(5))
x = 6/sqrt(5) = (6/5)sqrt(5)

Similarly, if y = -sqrt(5), then x = -(6/5)sqrt(5)

So our solution is
{x = (6/5)sqrt(5), y = sqrt(5)} or
{x = (-6/5)sqrt(5), y = -sqrt(5)}

Answer 3

The given equations are
y^2 +2xy - 16 =0 and

3y^2 -xy-2 +0

Multiply the second by 2 and add it with first to get

7y^2 - 20 = 0.

y = + ( 20/7)^0.5 or - ( 20/7) ^0.5

Use these values in any one of the given two equations to find the values of x.
x = - 3.89 or +3.89.

Answer 4

2xy + y^2 = 16
-xy + 3y^2 = 2

Eliminate xy .... by multiplying the second eqn (both sides) with 2 and adding the same to the first equation ....
7y^2 = 20
y = Sqrt (20/7) Then substitute this any one eqn to get x

Answer 5

2xy + y² = 16 .... (1)
3y² - xy = 2 ........(2)

Equation (1) + 2 * equation (2)

6y² + y² = 20

7y² = 20
y² = 20/7
y = ±√(20/7)

Substitute into equation (1)

2x * ±√(20/7) + 20/7 = 16

So ±2√(20/7) * x = 92/7

So x = ±(46/7)/√(20/7)
= ±(46/7 * 7/20)* √(20/7)
= ±2.3√(20/7)

Check in equation (2)
3y^2 - xy = 3 * 20/7 - 2.3*20/7
= 60/7 - 46/7
= 14/7
= 2 YESSSSSSSSSSSSSSSSSSSSSSSS!!!!!!!!!!!!!!!!!!!!

So the solution is x = 2.3√(20/7), y = √(20/7)
OR x = -2.3√(20/7), y = -√(20/7)

Answer 6

since you are dealing with y^2 and xy only, you could conceivably set up a 2x2 matrix and solve by Gaussian elimination.

Otherwise use Newton's method. Calculate the Jacobian of the matrix, and start with an initial guess of, say, (x,y)=(0,0) and Newton's method should get you a decent answer within 2-3 steps.