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∙ 9y agox^2 + y^2 = 9 eq(1)
x + y = 2 eq (2)
thus
y = 2-x
substitute into (1)
x^2 + (2-x)^2 = 9
x^2 +4 -4x +x^2 = 9
2x^2 -4x -5 = 0
x = (-b +/- root(4ac)) / 2
x =( 4 + 7.483) / 4 = 2.87
y = -0.87
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∙ 9y agoWiki User
∙ 9y agox2 + y2 = 9 ..................... 1x + y = 2 ......................... 2
from 2, y = 2 - x ..............3
Substitute in 1,
x2 + (2-x)2 = 9
x2 + 4 + x2 -4x = 9
2x2 -4x - 5 = 0
Solve for x and then substitute in 3 for y.
They are: (3, 1) and (-11/5, -8/5)
The solutions are: x = 4, y = 2 and x = -4, y = -2
Simultaneous equations.
The system is simultaneous linear equations
Another straight line equation is needed such that both simultaneous equations will intersect at one point.
They are: (3, 1) and (-11/5, -8/5)
The solutions are: x = 4, y = 2 and x = -4, y = -2
Simultaneous suggests at least two equations.
If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)
Simultaneous equations.
The system is simultaneous linear equations
x = -3 y = -2
Another straight line equation is needed such that both simultaneous equations will intersect at one point.
Merge the equations together and form a quadratic equation in terms of x:- 3x2-20x+28 = 0 (3x-14)(x-2) = 0 x = 14/3 or x = 2 So when x = 14/3 then y = -13/3 and when x = 2 then y = 1
1st equation: x^2 -xy -y squared = -11 2nd equation: 2x+y = 1 Combining the the two equations together gives: -x^2 +3x +10 = 0 Solving the above quadratic equation: x = 5 or x = -2 Solutions by substitution: (5, -9) and (-2, 5)
The two rational solutions are (0,0,0) and (1,1,1). There are no other real solutions.
They are two equations in two unknown variables (x and y), which are inconsistent. That is to say, there is no simultaneous solution. An alternative approach is to say that they are the equations of two lines in the Cartesian plane. The lines are parallel and so they do not meet indicating that there is no simultaneous solution.