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What are the points of intersection when the line x -y equals 2 crosses the curve x squared - 4y squared equals 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)
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What are the points of intersection when the line y equals 10x -12 crosses the curve y equals x squared plus 20x plus 12?
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)
What are the points of intersection of the parabolas of y equals 4x squared -12x -3 and y equals x squared plus 11x plus 5 on the Cartesian plane?
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)
Why can three coplanar lines have zero one two or three points of intersection?
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.
What are the points of intersection of the parabolas y equals x squared -2x plus 4 and y equals 2x squared -4x plus 4?
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)
What are the points of intersection of the line x -2y equals 1 with the curve 3xy -y squared equals 8 showing work?
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)
What are the points of intersection of the line x -y equals 2 with x squared -4y squared equals 5?
The points of intersection are: (7/3, 1/3) and (3, 1)
What are the points of intersection of 3x -2y -1 equals 0 with 3x squared -2y squared equals -5 on the Cartesian plane?
Points of intersection work out as: (3, 4) and (-1, -2)
Where are the points of intersection of the equations 4y squared -3x squared equals 1 and x -2y equals 1?
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
What are points of intersection of the line 3x -y equals 5 with the curve of 2x squared plus y squared equals 129?
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)
What are the points of intersection of the line y equals -8-3x with the curve y equals -2-4x-x squared?
They work out as: (-3, 1) and (2, -14)
What are the points of intersection when the line y equals 10x -12 crosses the curve y equals x squared plus 20x plus 12?
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)
What are the point of intersection of x is equal to y squared plus 2y?
You need two, or more, curves for points of intersection.
What are the points of intersection of y equals x squared plus 2x plus 2 with y equals x cubed plus 2?
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)
What are the points of intersection of the parabolas of y equals 4x squared -12x -3 and y equals x squared plus 11x plus 5 on the Cartesian plane?
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)
What are the points of intersection when the line 3x -y equals 5 crosses the curve 2x squared plus y squared equals 129?
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)
What are the points of intersection of x plus y equals 7 with x squared plus y squared equals 25 showing key stages of work?
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)
Where are the points of contact when the line 2x plus 5 equals 4 crosses the curve y squared equals x plus 4 on the Cartesian plane?
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
