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)
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
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Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.There are no equals signs and so no equations to plot!Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.There are no equals signs and so no equations to plot!Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.There are no equals signs and so no equations to plot!Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.There are no equals signs and so no equations to plot!
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
The points of intersection are: (7/3, 1/3) and (3, 1)
Points of intersection work out as: (3, 4) and (-1, -2)
You need two, or more, curves for points of intersection.
No, they do not.
No, linear equations don't have x2. Equation with x and y are usually linear equations. Equations with either x2 or y2 (but never both) are usually quadratic equations.
Times itself. For example, 4 squared would be 4x4 = 16. 5 squared would be 5x5 = 25 etc
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These are two expressions, not equations. Expressions do not have solutions, only equations do. NB equations include the equals sign.
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)