If: 2x+y = 1 then y = 1-2x and y^2 = (1-2x)^2 which is 1-4x+4x^2
If: x^2 -xy -y^2 = -11 then x^2 -x(1-2x) -(1-4x+4x^2) = -11
So it follows: x^2 -x +2x^2 -1 +4x -4x^2 +11 = 0
Collecting like terms: -x^2 +3x +10 = 0
Factorizing the above: (x-5)(-x-2) = 0 meaning x = 5 or -2
Points of intersection by substitution are at: (5, -9) and (-2, 5)
They are (0, 1) and (3, -5).
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)
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.
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)
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)
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 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)
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)
If: y = x^2 +2x +2 and y = 7 -2x Then: x^2 +2x +2 = 7 -2x Transposing terms: x^2 +4x -5 = 0 Factorizing the above: (x +5)(x -1) = 0 meaning x = -5 or x = 1 By substitution into original equations points of intersection are at: (-5, 17) and (1, 5)
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.
That system of equations has no solution. When the two equations are graphed, they turn out to be the same straight line, so there's no such thing as a single point where the two lines intersect. There are an infinite number of points that satisfy both equations.
Solving the simultaneous equations works out as x = -2 and y = -2 So the lines intersect at: (-2, -2)
The intersection is (-2, 6)