What is the value of k for which the system of equations 2x y 3 is equal to zero and 5 x KY 7 is equal to zero has no solution?

CBSE 10 - Maths

Asked by aayeshashaikh40.10 | 24 Jul, 2022, 06:42: PM

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.