If: 3x-5y = 16 then -5y = 16-3x => y = -3.2+0.6x by dividing all terme by -5
If: xy = 7 then x(-3.2+0.6x) = 7 => -3.2x+0.6x2 -7 = 0
Solving the quadratic equation: x = 7 or x = -5/3
Solutions of the equations by substitution are: x = 7, y = 1 and x = -5/3, y = -21/5
Through a process of elimination and substitution the solutions are s = 8 and x = 5
They are: (3, 1) and (-11/5, -8/5)
They are simultaneous equations and their solutions are x = 41 and y = -58
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