So by definition both equations equal one another:
5x2-2x+1 = 6-3x-x2
5x2+x2-2x+3x+1-6 = 0
6x2+x-5 = 0
Factor the above equation with the help of the quadratic equation formula:
(6x-5)(x+1) = 0
When x = -1 y = 8
When x = 5/6 y = 101/36
Therefore the curves intersect at points: (-1, 8) and (5/6, 101/36)
You need two, or more, curves for points of intersection.
The points of intersection are: (7/3, 1/3) and (3, 1)
Points of intersection work out as: (3, 4) and (-1, -2)
The points of intersection. The coordinates of such points will be the solutions to the simultaneous equations representing the curves.
They work out as: (-3, 1) and (2, -14)
You need two, or more, curves for points of intersection.
The points of intersection are: (7/3, 1/3) and (3, 1)
Points of intersection work out as: (3, 4) and (-1, -2)
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
The points of intersection. The coordinates of such points will be the solutions to the simultaneous equations representing the curves.
Straight line: 3x-y = 5 Curved parabola: 2x^2 +y^2 = 129 Points of intersection works out as: (52/11, 101/11) and (-2, -11)
They work out as: (-3, 1) and (2, -14)
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
The points are (-1/3, 5/3) and (8, 3).Another Answer:-The x coordinates work out as -1/3 and 8Substituting the x values into the equations the points are at (-1/3, 13/9) and (8, 157)
There are none. Those two curves do not intersect.
If: x+y = 7 and x2+y2 = 25 Then: x = 7-y and so (7-y)2+y2 = 25 => 2y2-14y+24 = 0 Solving the quadratic equation: y = 4 and y = 3 By substitution points of intersection: (3, 4) and (4, 3)
Solutions may be closed or open regions or they may be points within a region (for example, grid points for integer solutions), or points of intersection between curves or between curves and the axes. It all depends on what the graphs and the solutions are.