Question

For what value of k, the system of equations
kx + 2y = 5,
3x - 4y = 10

has (i) a unique solution, (ii) no solution?

Answer 1

The given system of equations:
kx + 2y = 5
⇒ kx + 2y - 5 = 0                ….(i)
3x - 4y = 10
⇒3x - 4y - 10 = 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 = k, b_1= 2, c_1= -5 and a_2 = 3, b_2= -4, c_2= -10`
(i) For a unique solution, we must have:
∴ `(a_1)/(a_2) ≠ (b_1)/(b_2)  i.e., k/3 ≠ 2/(−4) ⇒ k ≠ (−3)/2`
Thus for all real values of k other than `(−3)/2`, 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)`
`⇒ k/3 = 2/(−4) ≠ (−5)/(−10)`
`⇒ k/3 = 2/(−4) and k/3 ≠ 1/2`
`⇒k = (−3)/2, k ≠ 3/2`
Hence, the required value of k is `(−3)/2`.