Consider the system of equations y + 2kz = 0 x + 2y + 6z = 2 kx + 2z = 1 where k is an arbitrary constant. (a) For which values of the constant k does this system have a unique solution? (b) For which values of the constant k does this system have no solution?

the line of

step-by-step explanation: 4 cause 2+2=4

it’s the first one your welcome

step-by-step explanation:

step 1: solve one of the equations for one of the variables. let's solve the first equation for y:

step 2: substitute that equation into the other equation, and solve for x.

step 3: substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.