For what value of k does the pair of equations x + 2y=3 3x ky + 7 = 0 have a unique solution

Text Solution

Answer :  (i) `k ne 10 ` (ii) `k = 10`

Solution :  (i) ` (1)/(5) ne (2)/(k) rArr k ne 10 ` <br> (ii)` (1)/(5) = (2)/(k) ne (-3)/(7) rArr k = 10 ` <br> Clearly, ` (1)/(5) = (2)/(k) = (-3)/(7) ` is never true, as ` (1)/(5) = ( -3)/(7) ` is false.

The given system of equations:x + 2y = 3⇒ x + 2y - 3 = 0                           ….(i)And, 5x + ky + 7 = 0                   …(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 1, b_1= 2, c_1= -3 and a_2 = 5, b_2 = k, c_2 = 7`(i) For a unique solution, we must have:∴ `(a_1)/(a_2) ≠ (b_1)/(b_2) i.e., 1/5 ≠ 2/k ⇒ k ≠ 10`Thus for all real values of k other than 10, the given system of equations will have a unique solution.(ii) In order that the given system of equations has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2 )≠ (c_1)/(c_2)``⇒ 1/5 ≠ 2/k ≠ (−3)/7``⇒ 1/5 ≠ 2/k and 2/k ≠ (−3)/7``⇒k = 10, k ≠ 14/(−3)`Hence, the required value of k is 10.

There is no value of k for which the given system of equations has an infinite number of solutions.