# System of equations

1. May 29, 2010

### evagelos

1. The problem statement, all variables and given/known data

How do we prove the system of the following equations??

1) $$3xy+y^2 =-7$$

2) $$x^2-2xy = 30$$

2. Relevant equations

3. The attempt at a solution

I tried by solving the 1st equation in terms y and then substituting into equation (2) but the problem got more complicated

2. May 29, 2010

### rock.freak667

Can you please post your working, because substitution seems like the only viable method.

3. May 30, 2010

### evagelos

$$2y^2 +3xy +7 =0$$ and solving for y we have:

$$y= \frac{-3x+\sqrt{9x^2-56}}{4}$$ or $$y=\frac{-3x-\sqrt{9x^2-56}}{4}$$

Now we substitute for y in the 2nd equation and we get two equations:

$$x^2-2x\frac{-3x+\sqrt{9x^2-56}}{4}-30 =0$$

......................or..................................

$$x^2-2x\frac{-3x-\sqrt{9x^2-56}}{4}-30 =0$$

4. May 30, 2010

### ehild

This is the hardest way to proceed. First solve the first equation for x, or the second equation for y.

ehild

5. May 30, 2010

### evagelos

and then substitute in the 2nd equation??

6. May 30, 2010

### rock.freak667

Yes.

7. May 30, 2010

### iRaid

Solve by substitution, I'll give you the first 1 (MODS PLEASE DON'T INFRACT ME I'M NOT GIVING HIM THE FULL ANSWER!! < IF SOMETHING IS WRONG PLEASE PM ME NOT INFRACT :()

2(3xy+y2=-7)
3(-2xy+x2=30)
=
6xy+2y2=-14
-6xy+6y2=90
Then you'll be left with 1 variable :p

8. May 31, 2010

### Susanne217

Sorry Cause I'm silly gringa....

Anyway back to the original problem

if you try to solve this regulary you get som very silly fractions...

Last edited by a moderator: Jun 11, 2010
9. May 31, 2010

### Staff: Mentor

How is it related to the original problem?

10. May 31, 2010

### evagelos

Thank ,it works out.

11. May 31, 2010

### HallsofIvy

Well, only because you have changed $x^2$ to $y^2$ in the second equation!

You should have $6xy+ 2y^2= -14$ and $-6xy+ 6x^2= 90$. Adding those two equations gives you $6x^2+ 2y^2= 76$. You still have two variables.

12. Jun 1, 2010

### Susanne217

isn't there a mistake there? its $$y^2 +3xy +7 =0$$

13. Jun 11, 2010

### Susanne217

I have tried and tried for the last couple of weeks to try to solve this system using pre calculus tools and I always come up with that the two parts of the system do not intersect and there doesn't exist a commen set of fix points

I tried to draw a phase portrait in Maple and it does look like something interesting happens in and around the origin and that there asome asympotetes, but I can't get a userable solution :(

Maybe there is a mistake by the original poster?

14. Jun 11, 2010

### vela

Staff Emeritus
I plotted the original equations in Mathematica, and there are in fact four solutions.

15. Jun 12, 2010

### ehild

Solve the second equation for y:

$$y=\frac{x^2-30}{2x}$$

Substitute into the first equation and simplify:

$$3x\frac{x^2-30}{2x}+\frac{(x^2-30)^2}{4x^2}=-7$$

As x is not equal to 0, you can multiply the whole equation with 4x^2.

$$6x^2(x^2-30)+(x^2-30)^2=-28x^2$$

By simplifying again, you get

$$7x^4-212x^2+900=0$$

Solve for x^2.

$$x^2=\frac{106\pm \sqrt{4936}}{7}$$

x1=5.02, y1=-0.48
x2=-5.02, y2=0.48
x3=2.26, y3=-5.51
x4=-2.26, y4=5.51

ehild