x^2+4y^2=36

2y-x-6=0

Printable View

- July 24th 2008, 12:46 PMjimCsolving systems of equations with conics
x^2+4y^2=36

2y-x-6=0 - July 24th 2008, 01:20 PMTKHunny
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. (Clapping) - July 24th 2008, 01:54 PMjimCi want it done in elimination
elimination please

- July 24th 2008, 03:44 PMTKHunny
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

You can add.

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". - July 24th 2008, 03:53 PMjimCk i missed this class so i need all the work i cant exactly get them started.
keep going bud.

- July 24th 2008, 04:19 PMmr fantastic
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?