For which values of k does the system of linear equations have zero, one, or an infinite number of solutions? [Note: not all three possibilities need occur.] (If the answer is an interval of numbers, enter your answer using interval notation. If an answer does not exist, enter DNE.) 8x1 + x2 = 2 kx1 + 9x2 = 18

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.

in mathematics, a constraint is a condition of an optimization problem that the solution must satisfy. there are several types of constraints—primarily equality constraints, inequality constraints, and integer constraints. the set of candidate solutions that satisfy all constraints is called the feasible set.

hoped i'm right lol