Question

For what value of k, the system of equations
x + 2y = 3,
5x + ky + 7 = 0
Have (i) a unique solution, (ii) no solution?
Also, show that there is no value of k for which the given system of equation has infinitely namely solutions

Answer 1

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.