x^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
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
The two rational solutions are (0,0,0) and (1,1,1). There are no other real solutions.
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)
They are simultaneous equations and their solutions are x = 41 and y = -58