# Thread: solving systems of equations with conics

1. ## solving systems of equations with conics

x^2+4y^2=36
2y-x-6=0

2. Originally Posted by jimC
x^2+4y^2=36
2y-x-6=0
There is no reason to let this scare you.

You already know how to solve a linear equation.

2y-x-6=0

x = 2y - 6

You already now how to substitute.

x^2+4y^2=36

(2y - 6)^2+4y^2=36

Expanding and Simplifying are not new

4y^2 - 24y + 36 + 4y^2 = 36

8y^2 - 24y + 36 = 36

8y^2 - 24y = 0

y^2 - 3y = 0

If all else fails, we could here resort to the Quadratic Formula. In this case, since we lost the constant term, it factors easily.

y(y - 3) = 0

y = 0 or y = 3

My point here, is that you should not consider every problem you have not seen as a brand new experience. You have the tools. Use them.

3. ## i want it done in elimination

4. Originally Posted by jimC
x^2+4y^2=36
2y-x-6=0
There is no reason to let this scare you.

x^2+4y^2=36
2y-x-6=0

You can rearrange

x^2 + 4y^2 = 36
-x + 2y = 6

You can multiply (Second equation by x)

x^2 + 4y^2 = 36
-x^2 + 2yx = 6x

4*y^2 + 2xy = 36 + 6x

It still has the Quadratic Formula written all over it.

Again, nothing new. Again, you managed to show no work.

I'm not sure we're getting anywhere, but you're the one who insisted on "elimination".

5. ## k i missed this class so i need all the work i cant exactly get them started.

keep going bud.

6. Originally Posted by jimC
keep going bud.
I think TKH has gone as far as needed.

Why do you want elimination, anyway? The technique used in post #2 effectively is elimination since x got eliminated. As you can plainly see from post #4, the elimination method per se is not effective for this question.

You say you missed this class. Have you spoken to your teacher? Have you attempted to get a copy of the class notes from another student? Do you have a textbook with examples?