Best Answer

The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)

Q: Where are the points of intersection of the equations 4y squared -3x squared equals 1 and x -2y equals 1?

Write your answer...

Submit

Still have questions?

Continue Learning about Algebra

The points of intersection are normally the solutions of the equations for x and y

If: y = x^2 +3x -10 and y = -x^2 -8x -15 Then: x^2 +3x -10 = -x^2 -8x -15 Transposing terms: 2x^2 +11x +5 = 0 Factorizing the above: (2x +1)(x +5) = 0 meaning x = -1/2 or -5 Therefore by substitution points of intersection are at: (-1/2, -45/4) and (-5, 0)

The first graph consists of all points whose coordinates satisfy the first equation.The second graph consists of all points whose coordinates satisfy the second equation.The point of intersection lies on both lines so the coordinates of that poin must satisfy both equations.

it will form a parabola on the graph with the vertex at point (0,0) and points at (1,1), (-1,1), (2,4), (-2,4)......

A: 3x-2y = 1 => 3x = 1+2y B: 3x2-2y2+5 = 0 => 3x2 = 2y2-5 Square both sides in equation A: 9x2 = 1+4y+4y2 Multiply all terms by 3 in equation B: 9x2 = 6y2-15 So it follows that:- 6y2-15 = 1+4y+4y2 and 6y2-15-1-4y-4y2 = 0 Collect like terms: 2y2-4y-16 = 0 Divide all terms by 2: y2-2y-8 = 0 Factorise: (y-4)(y+2) = 0 Therefore: y = 4 or y = -2 Substitute the above y values into the linear equation to find the values of x: The points of intersection are: (3, 4) and (-1,-2)

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 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 are normally the solutions of the equations for x and y

If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 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)

We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.

It works out that the points of intersection between the equations of 2x+5 = 5 and x^2 -y^2 = 3 are at: (14/3, -13/3) and (2, 1)

Equations: 3x-2y = 1 and 3x^2 -2y^2 +5 = 0 By combining the equations into one single quadratic equation and solving it the points of contact are made at (3, 4) and (-1, -2)

Equations: x -y = 2 and x^2 -4y^2 = 5 By combining the equations into a single quadratic equation in terms of y and solving it: y = 1/3 or y = 1 By means of substitution the points of intersection are at: (7/3, 1/3) and (3, 1)

The points of intersection. The coordinates of such points will be the solutions to the simultaneous equations representing the curves.