The system has 8x 4y 2 20 x ky 1 has a unique solution

The system has -8x + 4y = -2 -20 x + ky = - 1 has a unique solution if k The solution is x =1 Your answers should depend on the value of k. Click here for more information about how to enter such expressions.

Solution

- 8x + 4y = - 2 or, 4x - 2y = 1... (!)

- 20 x + ky = -1 or, 20x - ky = 1...(2)

On multiplying both the sides of the 1st equation by 5, we get 20x - 10 y = 5... (3)

Now, on subtracting the 2nd equation from the 3rd equation, we get, 20x -10y - 20x + ky = 5 -1 or (k - 10) y = 4

Therefore, y = 4/ (k - 10). If k = 10, then since the denominator becomes 0 and since division by 0 is not defined, the given system will not have a unique solution. Also, when k= 10, the 2nd equation becomes, 20x -10y = 1 . Since the 3rd equation is 20x -10y = 5, we have 1 = 5 which is not correct. Thus, if k = 10, the given equations are inconsistent. The given system has a solution if k is not = 10.

If k is not = 10, then we have already determined that y = 4/ (k - 10). Also, from the 1st equation, we have 4x - 2y = 1. Therefore, 4x = 1 + 2y = 1 + [2*4/(k -10)] = 1 + 8/(k-10) =( k -10 +8) / ( k - 10) = (k +2)/ (k -10)

Thus, the solution is x = (k+2)/(k-10) and y = 4/ (k - 10)