If: y = 5x^2 -2x+1 and y = 6-3x-x^2
Then: 5x^2 -2x+1 = 6-3x-x^2
Transposing terms: 6x^2 +x -5 = 0
Factorizing the above: (6x-5)(x+1) = 0 meaning x = 5/6 or x = -1
By substituting the values of x into original equations points of intersection are at (-1, 8) and (5/6, 101/36)
The points of intersection are: (7/3, 1/3) and (3, 1)
Points of intersection work out as: (3, 4) and (-1, -2)
If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)
They work out as: (-3, 1) and (2, -14)
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
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 are: (7/3, 1/3) and (3, 1)
Points of intersection work out as: (3, 4) and (-1, -2)
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)
The points of intersection are normally the solutions of the equations for x and y
If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)
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)
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
It works out that the points of intersection between the equations of 2x+5 = 5 and x^2 -y^2 = 3 are at: (14/3, -13/3) and (2, 1)
Equations: 3x-2y = 1 and 3x^2 -2y^2 +5 = 0 By combining the equations into one single quadratic equation and solving it the points of contact are made at (3, 4) and (-1, -2)
Equations: x -y = 2 and x^2 -4y^2 = 5 By combining the equations into a single quadratic equation in terms of y and solving it: y = 1/3 or y = 1 By means of substitution the points of intersection are at: (7/3, 1/3) and (3, 1)