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You have a hyperbola xÂ² - 4yÂ² = 5, and a straight line x-y = 2. I've included a link to Wolfram Alpha, below where you can see the plot of these. We can solve this by substitution. Rearrange the equation of the line to be x = y + 2, then substitute this in the other equation:

(y + 2)Â² - 4yÂ² = 5. Expand out: yÂ² + 4y + 4 - 4yÂ² = 5.

Subtract 5 from both sides and combine like terms:

-3yÂ² + 4y - 1=0. Using the quadratic formula, we get y = (-4 +- sqrt(16 - 12))/(-6), so we have (-6)/(-6) = 1, or (-2)/(-6) = 1/3.

Take these two y values and plug in to the equation of the straight line, to get the x coordinates: x = y + 2. So for y = 1, we get x = 3, and for y = 1/3 we get x = 7/3. The two points are (7/3,1/3) and (3,1). Look at the graph and we can visually see that these two points are correct.

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

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)

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)

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)

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)

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)

You need two, or more, curves for points of intersection.

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)

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)