x+y=-3
x^2+2y^2-12y-18=0
2006-11-28 16:56:28 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x+y=-3
x=-y-3
x -3 -2 -1 0 1 2 3
y 0 -1 -2 -3-4-5 -6
x^2+2y^2-12y-18=0
when y=-1
x^2+2+12-18=0
x^2-4=0
x^2=4
x=+/-2
so (-2,-1),(2,-1)
when y=3
x^2+18-36-18=0
x^2-36=0
x=+/-6
so(-6,3),(6,3)
points of intersection
(-2,-1) and (-6,3)
2006-11-28 20:53:35 · answer #1 · answered by raj 7 · 0⤊ 0⤋
This is not difficult
x= -(3+y)
put in 2nd equation
(y+3)^2 +2y^2-12y-18 = 0
or y^2 +6y +9 - 2y^2 -12y - 18 =0
or 3y^2 - 6y - 9 =0
y^2-y-3 = 0
this you can solve for y
y = 1/2+sqrt(1+12)/2 or 1/2-sqrt(13)/2
from this you can find x
there are 2 points
(-3.5-sqrt(13)/2 , .5+sqrt(13/2))
and (-3.5 + sqrt(13/2) , .5 - sqrt(13)/2
2006-11-29 00:58:25 · answer #2 · answered by Mein Hoon Na 7 · 1⤊ 0⤋
x=-y-3 plug in for x and then you factor to get points
2006-11-29 00:57:53 · answer #3 · answered by dird 2 · 0⤊ 0⤋
Because quadratic equation.
you will get two vales of x and y
2006-11-29 01:00:50 · answer #4 · answered by aminnyus 2 · 0⤊ 0⤋
I don't know. What's da point man?
2006-11-29 00:57:42 · answer #5 · answered by Anonymous · 0⤊ 0⤋