substitution method?

Question

okay hahaha last question. I have a test tomorrow so I'm trying to go over my quizzes to study. Anyways. I'm a little stuck on two of the substitution problems. the first one is just a checking problem.

I've been graphing the 2 equations to make sure I got the right answers, but on the one that is:

x^2 -xy + y^2 =13
I can't remember how to get y-alone. I had this problem before but I'm just not seeing it. But I only need that so I can graph it and check my answers. the second problem is similar

x-2y = 4
x^2 -y^2 =11

I tried get x by itself in the first one but I didn't want to have to square a fraction like that when I substitute it back into the second problem. So I solved for y in the first one and substituted it into the second one. well, I ended up with

5y^2 + 16y + 5 =0
I'm wondering if I did that wrong because it won't factor and I really don't want to use the quadratic formula. I had a negative number under the square root last time. thanx!!!!

Answer 1

x^2 -xy + y^2 =13
you can only plot this
if you have the other equation . . . you can use the other equation to substitute to the first

x-2y = 4 . . . . . . . . equation 1 . . . x = 2y + 4 . .. substitute to eq.2
x^2 -y^2 =11 . . . . . equation 2
(2y + 4) ² - y ² = 11
4 y ² + 16 y + 16 - y ² = 11
3 y ² + 16 y + 5 = 0
(3 y + 1) (y + 5) = 0
y = - 1/3 . . . . and . . . y = - 5

when y = - 1/3
x = 2(-1/3) + 4 = (-2 + 12) /3 = 10 / 3

when y = -5
x = 2(-5) + 4 = - 6

Answer 2

I would use the quadratic formula to solve for y alone:

y² -xy + x²-13 = 0

y = [-(-x) ± sqrt(x² - 4(x²-13))]/2 = [x ± sqrt(52 - 3x²)]/2

The graph of y² -xy + x²-13 = 0 is an ellipse; each of the two "solutions" for y gives part of the ellipse.

For your second question, you were on the right track; you just made a small arithmetic error. The result of your substitution should be 3y² + 16y + 5 = 0.