1st equation: 5x +7y +29 = 0 or 5x +7y = -29
2nd equation: 11x -3y -65 = 0 or 11x -3y = 65
Times all terms in the 1st equation by 11: 55x +77y = -319
Times all terms in 2nd equation by 5: 55x -15y = 325
Subtract the 2nd equation from the 1st equation: 92y = -644 or y = -7
By means of substitution point of intersection is made at: (4, -7)
When x = -2 then y = 4 which is the common point of intersection of the two equations.
There are two equations in the question, not one. They are the equations of intersected lines, and their point of intersection is their common solution.
The intersection is (-2, 6)
Solving the simultaneous equations works out as x = -2 and y = -2 So the lines intersect at: (-2, -2)
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
When x = -2 then y = 4 which is the common point of intersection of the two equations.
There are two equations in the question, not one. They are the equations of intersected lines, and their point of intersection is their common solution.
x + y = 6x + y = 2These two equations have no common point (solution).If we graph both equations, we'll find that each one is a straight line.The lines are parallel, and have no intersection point.
The point of intersection of the given simultaneous equations of y = 4x-1 and 3y-8x+2 = 0 is at (0.25, 0) solved by means of elimination and substitution.
The intersection is (-2, 6)
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
Solving the simultaneous equations works out as x = -2 and y = -2 So the lines intersect at: (-2, -2)
It works out that the point of intersection is at (-4, -3.5) on the Cartesian plane.
2
The coordinates of the point of intersection is (1,1).
(4, -7)
It works out that they intersect at: (4, -7)