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.

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2