4x + 3y = -5 (Eq 1); 3x + 5y = -12 (Eq 2) Eliminate x Eq 1 times 3: 12x + 9y = -15 Eq 2 times 4: 12x +20y = -48 Subtract Eq 1 from Eq 2 11y = -48-(-15) ie 11y = -33 y = -3 Substitute in Eq 1: 4x - 9 = -5, ie 4x = 4 so x = 1 Check in Eq 2: (3 x 1) + (5 x -3) = 3 - 15 = -12. QED
16
You cannot solve 1 linear equation in 2 unknown variables. You need at least two independent equations.
(2,3)
It looks like you have 2 simultaneous equations with 2 variables:4x + 8y = 20 and -4x + 2y = -30. Solution is {x=7, y = -1}.One way to solve:Add the two equations together, combining like terms: (4x - 4x) + (8y + 2y) = 20-30 --> 0 + 10y = -10 --> y = -1. Substitute this into either of the original equations and solve for x=7, then check in the other equation to make sure you calculated correctly.
The solutions are: x = -2 and y = 4
Simultaneous equations.
The elimination method only works with simultaneous equations, hence another equation is needed here for it to be solvable.
4
Another straight line equation is needed such that both simultaneous equations will intersect at one point.
They are simultaneous equations and their solutions are x = 41 and y = -58
Through a process of elimination and substitution the solutions are s = 8 and x = 5
x = 4 and y = 0
You need three independent equations to solve for three unknown variables.
You can't solve it - you only have one equation and two unknowns. You need 2 equations to solve this.
You cannot solve this equation without some more information. The value of y depends upon the value of x, but they could take infinitely many different values. Maybe this is just one of a pair of simultaneous equations? If so, you need both equations to find values for x and y.
a=5: c=4
x = -1.2, y = -3