If: y = 5x^2 -2x +1 and y = 6 -3x -x^2
Then: 5x^2 -2x +1 = 6 -3x -x^2
Transposing terms: 6x^2 +x -5 = 0
Factorizing: (6x -5)(x +1) = 0 => x = -1 or x = 5/6
Through substitution points of intersect are at: (-1, 8) and (5/6, 101/36)
The points of intersection are: (7/3, 1/3) and (3, 1)
Points of intersection work out as: (3, 4) and (-1, -2)
They work out as: (-3, 1) and (2, -14)
If: y = 4x2-2x-1 and y = -2x2+3x+5 Then: 4x2-2x-1 = -2x2+3x+5 So: 6x2-5x-6 = 0 Solving the quadratic equation: x = -2/3 or x = 3/2 Points of intersection by substitution: (-2/3, 19/9) and (3/2, 5)
If: y = 4x^2 -2x -1 and 2x^2 = 3x -y +5 Then: 4x^2 -2x -1 = -2x^2 +3x +5 Transposing terms: 6x^2 -5x -6 = 0 Factorizing: (3x+2)(2x-3) = 0 => x = -2/3 or x = 3/2 By substitution points of intersection are at: (-2/3, 19/9) and (3/2, 5)
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 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)
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 = 4x2-2x-1 and y = -2x2+3x+5 Then: 4x2-2x-1 = -2x2+3x+5 So: 6x2-5x-6 = 0 Solving the quadratic equation: x = -2/3 or x = 3/2 Points of intersection by substitution: (-2/3, 19/9) and (3/2, 5)
If: y = 4x^2 -2x -1 and 2x^2 = 3x -y +5 Then: 4x^2 -2x -1 = -2x^2 +3x +5 Transposing terms: 6x^2 -5x -6 = 0 Factorizing: (3x+2)(2x-3) = 0 => x = -2/3 or x = 3/2 By substitution points of intersection are at: (-2/3, 19/9) and (3/2, 5)
If: y = 4x^2 -2x -1 and y = -2x^2 +3x +5 Then: 4x^2-2x-1 = -2x^2+3x+5 =>6x^2-5x-6 = 0 Solving the above quadratic equation: x = -2/3 or x = 3/2 Therefore by substitution the points of intersection are: (-2/3, 19/9) and (3/2, 5)
If: y = 4x^2 -2x -1 and y = -2x^2+3x+5 Then: 4x^2 -2x -1 = -2x^2+3x+5 => 6x^2-5x-6 = 0 Solving the above quadratic equation: x = -2/3 or x = 3/2 Therefore the points of intersection by substitution are: (-2/3, 19/9) and (3/2, 5)
If: y = -2x^2 +3x +5 and y = 4x^2 -2x -1 Then: 4x^2 -2x -1 = -2x^2 +3x +5 So it follows: 6x^2 -5x -6 = 0 Using the quadratic equation formula: x = -2/3 or x = 3/2 Therefore points of intersection by substitution are at: (-2/3, 19/9) and (3/2, 5)
If: y = x^2 -2x +4 and y = 2x^2 -4x +4 Then: 2x^2 -4x +4 = x^2 -2x +4 Transposing terms: x^2 -2x = 0 Factorizing: (x-2)(x+0) => x = 2 or x = 0 Therefore by substitution points of intersect are at: (2, 4) and (0, 4)