Best Answer

If: 2x +5y = 4

Then: x = -2.5y +2

If: y^2 = x +4

Then: x = y^2 -4

Therefore: y^2 -4 = -2.5y +2

Transposing the terms: y^2 +2.5y -6 = 0

Using the quadratic equation formula: y = -4 or y = 3/2

By substitution the line intersects the curve at: (12, -4) and (-7/4, 3/2)

Q: What are the points of intersection when the line 2x plus 5y equals 4 crosses the curve y squared equals x plus 4 showing work?

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

Points of intersection work out as: (3, 4) and (-1, -2)

They intersect at the point of: (-3/2, 11/4)

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)

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

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)

They intersect at the point of: (-3/2, 11/4)

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)

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)

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: 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)

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)