Solving a System of Multi-variable Polynomials: x^2+4y^2-8y=4, y^2-2y-8x-16=0

[NoahSilva2]

Solve the system of equations for x and y.

x²+4y²-8y=4
y²-2y-8x-16=0

This question was a bonus question I got lost on a test in my Algebra II Honors class.
So far, I've worked out that you need to subtract 4 from each side of the first equation to get the system of

x²+4y²-8y-4=0
y²-2y-8x-16=0

From here I know I need to factor, but we haven't learned yet how to factor quadrinomials with more than one variable. (Learned to factor by the box method)
Because the equations are equal to zero, I have a feeling there are two different solutions. I just dunno how to factor them.

Any help would be appreciated!

Thank you,
~NS2

 

[tkhunny]

There are two solutions. Solving simultaneous quadratics can lead to four!

In any case, you may find this WAY easier by drawing the pictures. Complete the Square in y for both and decide what sorts of things these equations represent. IT should be obvious from the pictures that there are two Real solutions.

From an algebra point of view, the easiest thing that pops out to me is the easy of solving the second equation for x.

From yet another perspective, take four times the second equation and subtract the result from the first equation. This will suggest strongly your two solutions, but it will be kidding a little, since one can't be valid. It DOES give a complete analysis of the x-value of the two solutions, but it will take some thought.

Truthfully, there are an awful lot of powers of 2 in there. I'd be tempted just to look around a little. Give x and y from 0, +/-2, +/-4, and maybe +/-8 a try and see where it leads.

 

[stapel]

Solve the system of equations for x and y.

x²+4y²-8y=4
y²-2y-8x-16=0

This question was a bonus question I got lost on a test in my Algebra II Honors class.
So far, I've worked out that you need to subtract 4 from each side of the first equation to get the system of

x²+4y²-8y-4=0
y²-2y-8x-16=0

From here I know I need to factor, but we haven't learned yet how to factor quadrinomials with more than one variable. (Learned to factor by the box method)

This doesn't matter. The factoring method is the same. However, it might be more useful to apply the Quadratic Formula, when dealing with this sort of messiness.

In the first equation, we have:

. . . . .\(\displaystyle x^2\, +\, 4(y^2\, -\, 2y)\, =\, 4\)

In the second equation, we have:

. . . . .\(\displaystyle (y^2\, -\, 2y)\, -\, 8x\, -\, 16\, =\, 0\)

The first equation gives us:

. . . . .\(\displaystyle y^2\, -\, 2y\, =\, \dfrac{4\, -\, x^2}{4}\)

Plugging this into the second equation, we get:

. . . . .\(\displaystyle \dfrac{4\, -\, x^2}{4}\, -\, 8x\, -\, 16\, =\, 0\)

. . . . .\(\displaystyle 4\, -\, x^2\, -\, 32x\, -\, 64\, =\, 0\)

. . . . .\(\displaystyle 0\, =\, x^2\, +\, 32x\, +\, 60\)

Where does this lead?