SOLUTION: two equations 5x^2 + 3y^2 =17 -x+y=-1 i need to find the points of intersection ,if any, of the graphs in the system

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: two equations 5x^2 + 3y^2 =17 -x+y=-1 i need to find the points of intersection ,if any, of the graphs in the system      Log On


   

Question 111011: two equations
5x^2 + 3y^2 =17
-x+y=-1
i need to find the points of intersection ,if any, of the graphs in the system
Found 2 solutions by checkley71, scott8148:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
5x^2+3y^2=17
3y^2=-5x^2+17 (red graph)
-x+y=-1
y=x-1 (green line)
+graph%28+300%2C+200%2C+-6%2C+5%2C+-10%2C+20%2C+3y%5E2=-5x%5E2%2B17%2C+y+=+x+-1%29+ (graph 300x200 pixels, x from -6 to 5, y from -10 to 20, of TWO functions 3y^2 = -5x^2 -17 and y = x -1).
y=x-1
3(x-1)^2=-5x^2+17
3(x^2-2x+1)=-5x^2+17
3x^2-6x+3=-5x^2+17
3x^2+5x^2-6x+3-17=0
8x^2-6x-14=0 reduce by dividing by 4
2(4x^2-3x-7)=0
2(4x-7)(x+1)=0
4x-7=o
4x=7
x=7/4 answer.
y=7/4-1
y=(7-4)/4
y=3/4 answer
so the intersection is (7/4,3/4)

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
from second equation y=x-1 ___ substituting into first equation gives 5x^2+3(x-1)^2=17

5x^2+3x^2-6x+3=17 ___ 8x^2-6x-14=0 ___ 4x^2-3x-7=0 ___ (4x-7)(x+1)=0

find the x values and plug into equation (the second one is easier) to find the corresponding y values for the (2) intersections