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
I suggest you solve the first equation for "x", replace that in the second equation, then use the quadratic formula to solve the combined formula. For each of the two values of "y" (you can usually expect two values for a quadratic equation), replace that in the first equation, to find the corresponding value for "x".
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.
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 system is simultaneous linear equations
The solutions work out as: x = 52/11, y = 101/11 and x = -2, y = -11
1 If: 2x+5y = 16 and -5x-2y = 2 2 Then: 2*(2x+5y =16) and 5*(-5x-2y = 2) is equvalent to the above equations 3 Thus: 4x+10y = 32 and -25x-10y = 10 4 Adding both equations: -21x = 42 or x = -2 5 Solutions by substitution: x = -2 and y = 4