If: 3x+2y = 5x+2y = 14
Then: 3x+2y = 14 and 5x+2y =14
Subtract the 1st equation from the 2nd equation: 2x = 0
Therefore by substitution the solutions are: x = 0 and y = 7
It has infinitely many solutions.
If: x-2y = 1 and 3xy -y^2 = 8 Then: x = 1+2y and 3(1+2y)y -y^2 = 8 Hence: 3y+6y^2 -y^2 = 8 => 5y^2 +3y -8 = 0 Solving the above quadratic equation: y =1 or y = -8/5 Solutions by substitution are: when y=1 then x=3 and when y=-8/5 then x=-11/5
It is a simultaneous equation and when solved its solutions are x = 71/26 and y = 50/13
The solutions work out as: x = 52/11, y = 101/11 and x = -2, y = -11
Simultaneous suggests at least two equations.
Equations: 7x-8y = 9 and 11x+3y = -17 Multiply all terms in the 1st equation by 11 and all terms in the 2nd by 7 So: 77x-88y = 99 and 77x+21y = -119 Subtract the 1st equation from the 2nd equation: 109y = -218 => y = -2 Through substitution the solutions are: x = -1 and y = -2
Another straight line equation is needed such that both simultaneous equations will intersect at one point.
The quadratic equation will have two solutions.
Through a process of elimination and substitution the solutions are s = 8 and x = 5
There are no solutions because the discriminant of this quadratic equation is less than zero
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)
using the t-table determine 3 solutions to this equation: y equals 2x
The solutions are: x = 4, y = 2 and x = -4, y = -2
They are: (3, 1) and (-11/5, -8/5)