SOLUTION: m^2 + 3n^2 = 11 m - n = 2 Conic - systems of two non-linear equations (two solutions) second question: y = x^2 and 7x = y + 12 (one solotion from these equations)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: m^2 + 3n^2 = 11 m - n = 2 Conic - systems of two non-linear equations (two solutions) second question: y = x^2 and 7x = y + 12 (one solotion from these equations)      Log On


   

Question 38194: m^2 + 3n^2 = 11
m - n = 2
Conic - systems of two non-linear equations (two solutions)

second question: y = x^2 and 7x = y + 12 (one solotion from these equations)
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
These are both solved the same way. We solve the linear equation and substitute for it into the curve...thus from
m^2 + 3n^2 = 11
m - n = 2
we have
m = n + 2
now substituting we get
(n + 2)^2 + 3n^2 = 11
n^2 + 4n + 4 + 3n^2 = 11
4n^2 + 4n - 7 = 0
This doesn't factor so use the quadratic and get
n = (-4 ± sqrt(128)) / 8
n = (-4 ± 8sqrt(2)) / 8
n = (-1 ± 2sqrt(2)) / 2
This gives you 2 values for n. To find each corresponding value of m, merely plug into m = n + 2 and you'll have your points of intersection.
The second problem is done the same way.