If: y = x2-7x+8 and y = -x2+9x-6
Then: x2-7x+8 = -x2+9x-6
So: 2x2-16x+14 = 0 => x2-8x+7 = 0
Therefore: x = 1 and x = 7
By substitution: x =1, y = 2 and x = 7, y = 8
Points of intersection: (1, 2) and (7, 8)
The intersection of the individual graphs. In the simplest case, the graph for each equation consists of a line (or some curve); the intersection is the points where the lines or curves meet.
If: x -2y = 1 then x = 1+2y If: 3xy -y^2 = 8 then 3(1+2y)y -y^2 = 8 So: 3y+6y^2 -y^2 = 8 => 3y+5y^2 -8 = 0 Solving the above quadratic equation: y = 1 or y = -8/5 By substitution the points of intersection are at: (3, 1) and (-11/5, -8/5)
geometry
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)
Geometry
The points of intersection. The coordinates of such points will be the solutions to the simultaneous equations representing the curves.
You need two, or more, curves for points of intersection.
The points of intersection are: (7/3, 1/3) and (3, 1)
There are none. Those two curves do not intersect.
Points of intersection work out as: (3, 4) and (-1, -2)
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.
The intersection of the individual graphs. In the simplest case, the graph for each equation consists of a line (or some curve); the intersection is the points where the lines or curves meet.
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
If: x-2y = 8 and xy = 24 Then: x = 8+2y and so (8+2y)y = 24 => 8y+2y2-24 = 0 Solving the quadratic equation: y = -6 or y = 2 By substitution points of intersection: (-4, -6) and (12, 2)
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)
If: x-2y = 1 and 3xy-y2 = 8 Then: x =1+2y and so 3(1+2y)y-y2 = 8 => 3y+5y2-8 = 0 Solving the quadratic equation: y = 1 or y = -8/5 Points of intersection by substitution: (3, 1) and (-11/5, -8/5)
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)