Answer:
For k = 36, there is 0 solutions
For other values of k, there is 1
We never got infinite solutions on the system.
Step-by-step explanation:
We have 2 unknowns and 2 equations:
E1 8x1 + x2 = 18
E2 k*22 x1 + 9 x2 = 18
If we multiply the E1 by 9 we obtain
E3 72 x1 + 9x2 = 162
if we substract E3 with E2 we obtain
(72- 2k) x1 = 144
Thus,
x1 = 144/ (72-2k)
That is, if 72-2k = 0, otherwise there is no solution. And 72-2k = 0 when k = 72/2 = 36.
If k is not 36, then
x1 = 144/(72-2k) and we can replace this value to obtain x2 by using E1
x2 = 18-8x1 = 18- 8 * (144/72-2k)
Which is a specific number that depends only on k. Thus,
for k = 36, there is 0 solutions
for other values of k, there is unique solution.
