Question

For what value of k, the system of equations
x + 2y = 5,
3x + ky + 15 = 0
has (i) a unique solution, (ii) no solution?

Answer 1

The given system of equations:
x + 2y = 5
⇒ x + 2y - 5 = 0            ….(i)
3x + ky + 15 = 0             …(ii) 
These equations are of the forms:
`a_1x+b_1y+c_1 = 0 and a_2x+b_2y+c_2 = 0`
where, `a_1 = 1, b_1= 2, c_1= -5 and a_2 = 3, b_2 = k, c_2 = 15`
(i) For a unique solution, we must have:
∴ `(a_1)/(a_2) ≠ (b_1)/(b_2) i.e., 1/3 ≠ 2/k ⇒ k ≠ `6`

Thus for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`
`⇒ 1/3 = 2/k≠ (−5)/15`
`⇒ 1/3 = 2/k and 2/k≠ (−5)/15`
⇒k = 6, k ≠ -6
Hence, the required value of k is 6.