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Q: What are the points of intersection when the line y equals 10x -12 crosses the curve y equals x squared plus 20x plus 12?

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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 all three lines are parallel, there are zero points of intersection. If all three lines go through a point, there is one point of intersection. If two lines are parallel and the third one crosses them, there are two. If the three lines make a triangle, there are three points.

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)

If: x -2y = 1 then x = 1+2y If: 3xy -y^2 = 8 then 3(1+2y)y -y^2 = 8 So: 3y+6y^2 -y^2 = 8 => 3y+5y^2 -8 = 0 Solving the above quadratic equation: y = 1 or y = -8/5 By substitution the points of intersection are at: (3, 1) and (-11/5, -8/5)

If: x -y = 2 then x^2 = y^2 +4y +4 If: x^2 -4y^2 = 5 then x^2 = 4y^2 +5 So: 4y^2 +5 = y^2 +4y +4 Transposing terms: 3y^2 -4y +1 = 0 Factorizing the above: (3y -1)(y -1) = 0 meaning y = 1/3 or y = 1 Substitution into the original linear equation intersections are at: (7/3, 1/3) and (3, 1)

Related questions

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)

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)

(3/4, 0) and (5/2, 0) Solved with the help of the quadratic equation formula.

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)

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)

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

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