Proving a System of Equations: 3xy+y^2=-7 and x^2-2xy=30

[evagelos]

Homework Statement

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

1) 3xy+y^2 =-7

2) x^2-2xy = 30

Homework Equations

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

 

[rock.freak667]

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

 

[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

 

[ehild]

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

ehild

 

[evagelos]

and then substitute in the 2nd equation??

 

[rock.freak667]

and then substitute in the 2nd equation??

Yes.

 

[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
Add those 2 together
Then you'll be left with 1 variable :p

 

[Susanne217]

1. Homework Statement

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

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...

 

[Borek]

something called the Jacobian Matrix

How is it related to the original problem?

 

[evagelos]

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

ehild

and then substitute in the 2nd equation??

Yes.

Thank ,it works out.

 

[HallsofIvy]

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
Add those 2 together
Then you'll be left with 1 variable :p

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.

 

[Susanne217]

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

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

 

[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?

 

[vela]

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

 

[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