Find the values of m and n such that the system has a no sol

Find the values of m and n such that the system has: (a) no solution; (b) infinitely many solutions; (c) a unique solution:

{ 12mx 3y + m = 0 and nx+2y=4

Solution

We are given

12mx -3y + m = 0...(1) and

nx + 2y = 4 or, nx + 2y - 4 = 0...(2) On multiplyoing 1st equation by 2 and the 2nd equation by 3, we get 24mx - 6y + 2m = 0...(3) and 3nx + 6y - 12 = 0...(4) On adding the 3rd and the 4th equations, we get 24mx - 6y + 2m + 3nx + 6y - 12 = 0 or, 24mx + 3nx = 12 - 2m or, (24m + 3n)x = 12 -2m or, x = (12 - 2m) / (24m + 3n). Now, since the division by 0 is not defined, if 24m + 3n = 0, i.e. if n = - 8m, the given system will not have any solution.

The given linear system has 2 equations in 2 variables. If these equations are identical, the system will have infinitely many solutions.From the 3rd and the 4th equations, we have 24mx - 6y + 2m = 0 and -3nx - 6y + 12 = 0 . If 2m = 12 and 3n = - 24m, i.e. if m = 6 and -3n = 24m or, n = - 8m = - 48, both of the 3rd and the 4th equation become, 144x - 6y + 12 = 0. This being a single equation in 2 variables, will have infinitely many solutions.

From the above, we already have x = (12 - 2m) / (24m + 3n). Let us assume that 24 m + 3n is not equal to 0. From the 2nd equation, we have nx + 2y - 4 = 0 or, 2y = 4 -nx = 4 - n(12 - 2m) / (24m + 3n) or, y = 2 - [(6 - m)/(24m + 3n)] . Thus if 24m + 3n 0, i.e. if n -8m, the given system will have a unique solution.