Hard math question having to do with systems of equations?

Question

I've been trying to figure out how to do this problem for days and I'm stuck. I've gone to friends and a teacher for help and I'm still stuck. I know what the answers are because I graphed it, but I can't figure out how to solve the problem algebraically. I've tried solving for x, for y and even x^2 but can never solve once I plug in the values. Can anyone help? All the students in my calculus class in college are having trouble with this problem, any ideas? (It's probly obvious but for some reason none of us r seeing how to do the problem.)-
Find all solutions to the system below algebraically:
2x^2 * y = 36
x^2 + y^2 = 13

Answer 1

To solve
2x^2 * y = 36
x^2 + y^2 = 13

Solve the first equation for x^2:
x^2 = 36/(2y) = 18/y

In order to replace x^2 in the second equation:
18/y + y^2 = 13

Multiply through by y:
18 + y^3 = 13y

Rearrange:
y^3 - 13y + 18 = 0

This is not an easy equation to solve. Since the product of the roots is -18, we can see if any factor of -18 is a root, and by trial and error, if nothing else, we can see that one such factor, 2, is indeed a root.

We can then divide this polynomial by y-2 and obtain a quadratic equation for the other two roots, which happen to be real.

There are various techniques for solving cubics more deterministically, but they are beyond the scope of a calculus class.

Answer 2

From the first equation, x^2 = 18/y; put this in the second to get 18/y+y^2=13, or 18+y^3=13y which is the cubic equation y^3-13y+18=0. There are method for solving cubics. There will be 3 values of y. Put these back into eq 1. There will be 2 values (+and-) for each y value.

Answer 3

from 1 x^2=18/y

plug into equation 2:

18/y+y^2=13 multiply by y

18+y^3=13y

f(y) y^3-13y+18=0

Now the above is cubic in y
in case this has got integral solution they are factor of 18 , +/-1, +/-2 etc
by trial and error 2 satisfies
f(2) = 8 - 26+18 =0
y^3-4y^-9y+18 =0
y(y^2-4) -9(y-2) =0
y(y+2)(y-2) -9(y-2) = 0
(y-2)(y^2+2y-9 = 0

y =2 is a solution so x = +/- 3
other solution is y is not rational can be solved

Answer 4

Taking into account eqn no.1 we find that y = 36/2x^2. For x = 0 there will be no solution for y. Now for x not equal to 0, we solve as below:

Substituting for X^2 from eqn. 2 in eqn 1 gives
2(13-y^2)y = 36
13y - y^3 = 18
y^3 - 13y + 18 = 0
(y-2)(y^2+2y-9) = 0

So the solution for y is y = 2 or y = (-2+Sq.Rt.40)/2 or
y= (-2-Sq.Rt.40)/2

i.e. y=2, y=-1 + 10^0.5, y= -1 - 10^0.5

Now substituting for each value of y, we get the values of x

For y = 2, from eqn 2 we get x^2 + 4 = 13
So, x^2 = 9. So the solution for x is x = 3 or x = -3

For y = -1+10^0.5, from eqn 2 we get,

x^2 + 11 - 2*10^0.5 = 13
x^2 = 2 + 2*10^0.5

So the solution for x is,
x = (2+2*10^0.5)^0.5 and x = - (2+2*10^0.5)^0.5

For y = -1 - 10^0.5, from eqn 2 we get,

x^2 + 11 + 2*10^0.5 = 13
x^2 = 2 - 2*10^0.5

So the solution for x is,
x = (2-2*10^0.5)^0.5 and x = - (2-2*10^0.5)^0.5

Hence the total solution for the equations in question are as below:

1. For x = 0, no solutions for y
2. y = 2, x = 3
3. y = 2, x = -3
4. y = -1 + 10^0.5, x = (2+2*10^0.5)^0.5
5. y = -1 + 10^0.5, x = -(2+2*10^0.5)^0.5
6. y = -1 - 10^0.5, x = (2-2*10^0.5)^0.5
7. y = -1 - 10^0.5, x = -(2-2*10^0.5)^0.5

Answer 5

(1) : 2x^2*y=36 becomes x^2=18/y
Substitute into equation (2)
18/y+y^2=13
y^3-13y+18=0
Using factor theorem we can find the first factor.
Let f(y)=y^3-13y+18
f(2)=8-26+18=0
y-2 is a factor of f(y)
y^3-13y+18=(y-2)(y^2+2y-9)=0
y=2 or y^2+2y-9=0
Use the quadratic equation formula to solve for the roots of
y^2+2y-9=0

Answer 6

x^2 + y^2 = 13
x^2 = 13 - y^2
2x^2*y = 36
2(13 - y^2)*y = 36
13y - y^3 = 18
y^3 - 13y +18 = 0
y^3 - 2y^2 + 2y^2 - 4y - 9y +18 = 0
y^2(y - 2) + 2y(y - 2) - 9(y - 2) = 0
(y - 2)(y^2 + 2y - 9) = 0
y - 2 = 0 or y^2 + 2y - 9 = 0
y = 2 or D = 4 + 36
y = 2 or = 40
y = 2 or y = (-2 + square root of 40)/2
y = 2 or y = -1 + square root of 10 or y = -1 - squre root of 40
x^2 = 13 - 4
x^2 = 9
x = 3 or x = -3

Answer 7

rewrite 1) as y = 36/(2x^2); for x not 0
substitute in second one et voila your eqation toolve

x^2 + (36/(2x^2))^2 = 13

or 4x^6 + 36^2 -52x^4 = 0
solve this for x and ur done, fot instamnce with the formulas of cardano.

for x = 0 there is no solution bcuz of 1)


ah better is to rewrite 1) as x^2 = 18/y abn substituet this in 2)
..

Answer 8

2x^2(y)=36
2x^2=36/y
x^2=18/y

plug into bottom equation:

18/y+y^2=13
multiply by y

18+y^3=13y
y^3-13y+18=0

If you know the roots then plug one into the above equation and then use synthetic or reqular division to reduce to a quadradic and then use the quad formula.

Answer 9

a million. x - 2y = -10 2. 3x - y = 0 for remedy structures of eq, we use between the eq to jot down one parameter consistent with yet another one: in this occasion we use 2, 3x - y = 0 -> y=3x After that, we use final word in yet another one eq, thus, a million: x - 2y = -10 -> x - 2*(3x) = -10 -> x - 6x = -10 -> -5x = -10 -> x = -10/(-5) = 2 .-> x=2 We write before that, y = 3x so y = 3(2) = 6 We solved it: x=2 and y=6

Answer 10

2x^2*y=36
x^2=36/2y
x=6sqrt(1/2y)
substitute this in
x^2+y^2=13
this gives
36*1/2y+y^2=13
and then solve this
you will get your result
thanks