Answer:
k ≠ 3
Step-by-step explanation:
Given the system of equation;
kx - y = 2 ------------------- 1
6x - 2y = 3 -------------------- 2
Rewriting the equations in the format ax+by+c = 0
Equation 1 becomes kx - y - 2 = 0
Equation 2 becomes 6x - 2y - 3 = 0
where a₁ = k, b₁ = -1 and c₁ = -2 and a₂ = 6, b₂ = -2 and c₂ = -3
For the system of equation to have a unique solution the following must be true;
a₁/a₂ ≠ b₁/b₁
Substituting the coefficients into the condition, we will have;
k/6 ≠ -1/-2
k/6 ≠ 1/2
Cross multiplying we will have;
2k ≠ 6
k ≠ 6/2
k ≠ 3
This means that k can be any other real values except 3 for the system of equation to have a unique solution.
