Q: What is the point of intersection between points 2x-4 and x plus 1?

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You need two, or more, curves for points of intersection.

There are none. Those two curves do not intersect.

2

The intersection is (-2, 6)

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)

Related questions

You need two, or more, curves for points of intersection.

There are none. Those two curves do not intersect.

2

The intersection is (-2, 6)

It works out that the point of intersection is at (-4, -3.5) on the Cartesian plane.

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)

points 12

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.

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)

x-y = 1 => x = y+1 x2+y2 = 5 => (y+1)(x+1)+y2 = 5 2y2+2y-4 = 0 y = -2 or y = 1 So the points of intersection are: (-1, -2) and (2, 1)

Equation: y = 8x^2 -26x+15 Equation when factorized: y = (4x-3)(2x-5) When x = 0 then y = (0, 15) which is the point of intersection on the y axis When y = 0 then x = (3/4, 0) and (5/2, 0) which are the points of intersection on the x axis

You have to draw the x and y axes, probably from -5 to 5 on each Then plot the point (-4,-2) and the point (0,2) because the intercept is two. Then draw a line between the two points, and continue it either side!