If: 3x-y = 5
Then: y^2 = 9x^2 -30x +25
If: 2x^2 +y^2 = 129
Then: 2x^2 +9x^2 -30x +25 = 129
Transposing terms: 11x^2 -30x -104 = 0
Factorizing the above: (11x-52)(x+2) = 0 meaning x = 52/11 or x = -2
Therefore by substitution points of intersection are at: (52/11, 101/11) and (-2, -11)
They are (-2, -11) and (4.7272..., 9.1818...).
The points of intersection are: (7/3, 1/3) and (3, 1)
Points of intersection work out as: (3, 4) and (-1, -2)
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
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: y = 10x -12 and y = x^2 +20x +12 Then: x^2 +20x +12 = 10x -12 Transposing terms: x^2 +10x +24 = 0 Factorizing: (x+6)(x+4) = 0 => x = -6 or x = -4 Points of intersection by substitution are at: (-6, -72) and (-4, -52)
You need two, or more, curves for points of intersection.
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)
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)
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)
(3/4, 0) and (5/2, 0) Solved with the help of the quadratic equation formula.
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))