Best Answer

Points of intersection work out as: (3, 4) and (-1, -2)

Q: What are the points of intersection of 3x -2y -1 equals 0 with 3x squared -2y squared equals -5 on the Cartesian plane?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

The points of intersection are: (7/3, 1/3) and (3, 1)

Improved Answer:-If: 2x+y = 5 and x^2 -y^2 = 3Then by rearranging: y = 5 -2x and -3x^2 -28+20x = 0Solving the above quadratic equation: x = 2 and x = 14/3By substitution points of intersection are: (2, 1) and (14/3, -13/3)

They work out as: (-3, 1) and (2, -14)

If: 3x-2y = 1 and 3x^2-2y^2+5 = 0 Then by rearranging: 3x = 2y+1 and y^2-2y-8 = 0 Solving the above quadratic equation: y = 4 or y = -2 Bu substitution points of intersection are: (3, 4) and (-1, -2)

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

Related questions

The points of intersection are: (7/3, 1/3) and (3, 1)

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 of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)

Improved Answer:-If: 2x+y = 5 and x^2 -y^2 = 3Then by rearranging: y = 5 -2x and -3x^2 -28+20x = 0Solving the above quadratic equation: x = 2 and x = 14/3By substitution points of intersection are: (2, 1) and (14/3, -13/3)

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)

If: 3x-2y = 1 and 3x^2-2y^2+5 = 0 Then by rearranging: 3x = 2y+1 and y^2-2y-8 = 0 Solving the above quadratic equation: y = 4 or y = -2 Bu substitution points of intersection are: (3, 4) and (-1, -2)

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

If: y = x^2 +3x -10 and y = -x^2 -8x -15 Then: x^2 +3x -10 = -x^2 -8x -15 Transposing terms: 2x^2 +11x +5 = 0 Factorizing the above: (2x +1)(x +5) = 0 meaning x = -1/2 or -5 Therefore by substitution points of intersection are at: (-1/2, -45/4) and (-5, 0)

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 = x2-x-12 Then points of contact are at: (0, -12), (4, 0) and (-3, 0)

The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))