I've seen multiple ways to solve systems of nonlinear equations, in this case ellipses, but the assignment I have gives me the steps to solve it. It says to solve for y in the first equation using the quadratic formula, which would leave me with a y= +/- and a radical. It then says to substitute that into the second equation, to collect like terms, then simplify. The following step says to move the remaining radical to the right side, then square the whole thing.

I can't get past how to get rid of the +/- algebraically. How are you supposed get rid of that without splitting it into two equations. But, according to my teacher that isn't the solution.

Two ellipses:
5x^2+y^2+30x-6y=-9
x^2+3y^2+4x-6y=21

Equation 1 after using quadratic formula:
y=3+/-SQRT(-5x^2-30x)

I just need help getting this one started,

2. Originally Posted by jsmith90210
...

Two ellipses:
5x^2+y^2+30x-6y=-9
x^2+3y^2+4x-6y=21

...
1. Get rid of the x²:

$\begin{array}{lcr}5x^2+y^2+30x-6y&=&-9\\
x^2+3y^2+4x-6y&=&21 \left| \cdot 5\right.\end{array}$

$\begin{array}{lcr}5x^2+y^2+30x-6y&=&-9\\
5x^2+15y^2+20x-30y&=&105 \end{array}$

Now subtract both equations columnwise:

$14y^2-10x-24y = 114~\implies~x=\dfrac1{10}\left(14y^2-24y-114\right)$

2. Plug in this value into one of the given equations. You'll get get an equation of degree 4. I found (by trial and error) 2 integer solutions. After long division I had to handle a quadratic equation.

3. Sorry

I apologize, but, the assignment was extremely specific, and in order to get credit, I have to do it as I laid out in the first post.

4. Originally Posted by jsmith90210
I apologize, but, the assignment was extremely specific, and in order to get credit, I have to do it as I laid out in the first post.
In that case, the work is meant to be your own work and not the work of others. Thread closed.