solve these simultaneous equations

Framed10

solve X^2 - xy + 3y^2 = 15 and x^2 + xy - y^2 = 5
have been messing around with solving these and no luck, am I missing something obvious?

Michael Collins

What have you got so far?

There are 4 answers, corresponding to the four intersection points (blue is first equation, red is second)

Raycyst

wow, very nice graph.

Both equations have a similar form and yet one is closed and one is open. Can you tell from the form of the equation what they'll look like when graphed?



I've being trying it in my head!
and in writing now in the reply box

If I multiple the second equation by three and add it to the first in order to eliminate y squared I seem to get something like
4X^2 + 2xy = 30.
X^2 + xy = 15
x(x+y)=15
hmmm.

edit.
more errors. Last lines there should be.
4X^2 + 2xy = 30. (now divide across by 2)
2X^2 + xy = 15
x(2x+y)=15
end edit.


I reckon I could subtract the equations to eliminate x squared.

4y^2 - 2xy = 10
2y^2 - xy = 5
y(2-x)=5
y=5/(2-x)

I'd now have to substitute for y in the first equation, and solve for x.
edit. Actually I've made a mistake there in the second part.
hmmm.

ok try again.

2y^2 - xy = 5
y(2y-x)=5
no good.
2y^2 - xy = 5
xy=2y^2 - 5
x=(2y^2 - 5)/y
and now substitute for x in the other equation and solve.

even now the question isn't near solved. From the pretty picture it seems you'd need four x values and four y values for the answer, and they probably need to be given as co-ordinate pairs.

These type questions weren't on honours leaving cert 25 years ago.

Michael Collins

Raycyst wrote: »

wow, very nice graph.

Both equations have a similar form and yet one is closed and one is open. Can you tell from the form of the equation what they'll look like when graphed?

Yep - they are both conic sections, i.e. a slice through a double cone.

The trick is to examine the x^2 and y^2 terms.

In the first equation, these have the same sign, but different coefficients => an ellipse (a slice fully through the cone, at an angle to the symmetry axis of a right cone).

In the second equation, these have opposite signs => therefore it is an hyperbola (the slice cuts through both halves of the double cone). An hyperbola has two branches, as we can see.

...even now the question isn't near solved. From the pretty picture it seems you'd need four x values and four y values for the answer, and they probably need to be given as co-ordinate pairs.

These type questions weren't on honours leaving cert 25 years ago.

Try eliminating the xy terms. This will allow you to obtain an expression for y^2 in terms of x^2, for example.

Now you can eliminate y from one of the original equations.

You'll end up with a 4th order equation, with no x^3 term, which can hence can be reduced to a quadratic in x^2.