If: x-2y+12 = 0 then x = 2y-12
If: x = 2y-12 then x^2 = (2y-12)^2 which is 4y^2-48y+144
If: x^2 +y^2 -x -31 = 0
Then: 4y^2 -48y +144+y^2 -(2y-12) -31 = 0
Removing brackets and collecting like terms: 5y^2 -50y +125 = 0
Divide all terms by 5 and factorize: (y-5)(y-5) = 0 => y = 5
Therefore by substitution solutions are: x = -2 and y = 5
They are simultaneous equations and their solutions are x = 41 and y = -58
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
Through a process of elimination and substitution the solutions are s = 8 and x = 5
They are simultaneous equations and their solutions are x = 41 and y = -58
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.
Do you mean: 4x+7y = 47 and 5x-4y = -5 Then the solutions to the simultaneous equations are: x = 3 and y = 5
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
I notice that the ratio of the y-coefficient to the x-coefficient is the same in both equations. I think that's enough to tell me that their graphs are parallel. So they don't intersect, and viewed as a pair of simultaneous equations, they have no solution.
The solutions work out as: x = 52/11, y = 101/11 and x = -2, y = -11
x = -3 y = -2