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

Q: What are the points of intersection of the line x -y equals 2 with x squared -4y squared equals 5?

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

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)

If: 3x-y = 5 and 2x2+y2 = 129 Then: 3x-y = 5 => y = 3x-5 And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0 Using the quadratic equation formula: x = 52/11 and x = -2 By substitution points of intersection are: (52/11, 101/11) and (-2, -11)

If: x-2y = 1 and 3xy-y2 = 8 Then: x =1+2y and so 3(1+2y)y-y2 = 8 => 3y+5y2-8 = 0 Solving the quadratic equation: y = 1 or y = -8/5 Points of intersection by substitution: (3, 1) and (-11/5, -8/5)

no 2 points form a line, 3 points form a plane

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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)

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)

If: 3x-y = 5 and 2x2+y2 = 129 Then: 3x-y = 5 => y = 3x-5 And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0 Using the quadratic equation formula: x = 52/11 and x = -2 By substitution points of intersection are: (52/11, 101/11) and (-2, -11)

intersection

YES. The intersection of two planes always makes a line. A line is at least two points.

If: x-2y = 1 and 3xy-y2 = 8 Then: x =1+2y and so 3(1+2y)y-y2 = 8 => 3y+5y2-8 = 0 Solving the quadratic equation: y = 1 or y = -8/5 Points of intersection by substitution: (3, 1) and (-11/5, -8/5)

no 2 points form a line, 3 points form a plane

If: 3x-y = 5 and 2x2+y2 = 129 Then: x = 5/3+y/3 and so 2(5/3+y/3)2+y2 = 129 => 50+20y+2y2+9y2 = 1161 Therefore: 11y2+20y-1111 = 0 Solving the quadratic equation: y = -11 or y = 101/11 By substitution points of intersection are: (-2, -11) and (52/11, 101/11)

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)

If: y = -8 -3x and y = -2 -4x -x^2 Then: -8 -3x = -2 -4x - x^2 Transposing terms: x^2 +x -6 = 0 Factorizing: (x-2)(x+3) = 0 => x = 2 or x = -3 Points of intersection by substitution are at: (2, -14) and (-3, 1)

x2+y2+4x+6y-40 = 0 and x = 10+y Substitute the second equation into the first equation: (10+y)2+y2+4(10+y)+6y-40 = 0 2y2+30y+100 = 0 Divide all terms by 2: y2+15y+50 = 0 (y+10)(y+5) = 0 => y = -10 or y = -5 Substitute the above values into the second equation to find the points of intersection: Points of intersection are: (0, -10) and (5, -5)