For which value(s) of k will the pair of equations kx + 3y = k – 3 and 12x + ky = k have no solution?

Step 2: By comparing the equations to the reference formula.

Using the conventional form to compare the two equations, we define

a1 = k, b1 = 3, and c1 = 3 – k

a2 = 12, b2 = k, and c2 = -k

Step 3: Write the circumstances in which there are no solutions.

The supplied set of equations must meet the condition if there is no solution.

a2/a2 = b2/b2 ≠ c2/c2

k/12 = 3/k ≠ (3-k)/-k

Step 4: Obtain value for k

From the first two parts,

k/12 = 3/k

k2 – 36 = 0

k = ± 6

From the last two parts,

3/k ≠ (3-k)/-k

k – 3 ≠ 3

k ≠ 6

From the last two results, we know that for k = -6, the equations will have no solutions

Hence, for k = -6, the set of equations has no solutions.

Summary:

For which value(s) of k will the pair of equations kx + 3y = k – 3 and 12x + ky = k have no solution?

For k = -6, the pair of equations kx + 3y = k – 3 and 12x + ky = k have no solution. The pair of equations has to be written in standard form and solved by comparison in order to obtain the k value.