Advanced Trig Help?

Question

For what value of k does the system of equations,
2x + y = 1; 12x + ky = 12 have no solution?

Please show work, explain, do whatever so that I can see how it is done.

Thanx a bunch!

Answer 1

- 12x - 6y = - 6
12x + ky = 12----ADD

(k - 6) y = 6
Has no solution if k = 6

Answer 2

What does this have to go with trigonometry? This is just algebra.

In order for this to have no solution, the 2x2 matrix of coefficients must have determinant 0. In other words,

[2 1 / 12 k] must have determinant 0 (the "/" separates rows).

Thus, we must have:

2k - 12 = 0 => k = 6,

for no solution to exist.

As you can see, now we have:

2x + y = 1
12x + 6y = 12

But then you have:

12 = 12x + 6y = 6 * (2x + y) = 6,

which is false.

Answer 3

when the two lines are parallel, they don't intersect, and that's why there is no solution. So basically, we are finding the value of k that makes the slope of the equation 12x + ky = 12 equals to the slope of the equation 2x + y = 1

solve for y
2x + y = 1
y = -2x + 1

m = -2

the slope of the equation 12x + ky = 12 must be also -2. Solve for y
12x + ky = 12
ky = -12x + 12
y = (-12/k)x + 12/k

the slope of the equation is -12/k, and we want it to equal to -2

-12/k = -2
-12 = -2k
k = 6

so the value of k is 6