Question

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

Answer 1

The given system of equations is:
x + 2y = 3
x + 2y − 3= 0 ....(i)
And, 5x + ky + 7 = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁= 2, c₁ = −3 and a₂ = 5, b₂ = k, c₂ = 7
(i) For a unique solution, we must have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, i.e. $\frac{1}{5}\ne \frac{2}{k}\Rightarrow k\ne 10$
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:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{1}{5}=\frac{2}{k}\ne \frac{-3}{7}$
$\Rightarrow \frac{1}{5}=\frac{2}{k}\mathrm{and}\frac{2}{k}\ne \frac{-3}{7}$
$\Rightarrow k=10,k\ne \frac{14}{-3}$
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.