Question

Find the value of k for which the system of equations 3x + y = 1 and kx + 2y = 5 has (i) a unique solution, (ii) no solution.

Answer 1

1. Identify the coefficients

Write the system of equations in the standard form a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

Equation 1:

kx + 2y = 5

⇒ a₁ = k, b₁ = 2, c₁ = 5

Equation 2:

3x + y = 1

⇒ a₂ = 3, b₂ = 1, c₂ = 1

2. Condition for unique solution

A linear system has a unique solution if the two lines intersect at exactly one point. This happens when the ratios of their coefficients are unequal:

`(a_1)/(a_2) ≠ (b_1)/(b_2)`

Substitute the values into the inequality:

`k/3 ≠ 2/1`

k ≠ 6

Thus, the system has a unique solution for all real numbers except k = 6.

3. Condition for no solution

The system has no solution if the two lines are parallel and never intersect. This occurs when the x and y coefficient ratios are equal, but different from the constant ratio:

`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`

Substitute the values into the equation:

`k/3 = 2/1 ≠ 5/1`

From the first part:

`k/3 = 2`

⇒ k = 6

Since 2 ≠ 5, the condition `(b_1)/(b_2) ≠ (c_1)/(c_2)` is satisfied. Thus, the system has no solution when k = 6.