Question

Find the value of k for which each of the following system of equations has no solution:
$3x+y=1,(2k-1)x+(k-1)y=(2k+1).$

Answer 1

The given system of equations:
3x + y = 1
⇒ 3x + y − 1 = 0 ...(i)
And, (2k − 1)x + (k − 1)y = (2k + 1)
⇒ (2k − 1)x + (k − 1)y − (2k + 1) = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 3, b₁= 1, c₁ = −1 and a₂ = (2k − 1), b₂ = (k − 1), c₂ = −(2k + 1)
In order that the given system of equations has no solution, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
i.e. $\frac{3}{\left(2k-1\right)}=\frac{1}{\left(k-1\right)}\ne \frac{-1}{-\left(2k+1\right)}$
$\frac{3}{\left(2k-1\right)}=\frac{1}{\left(k-1\right)}$ and $\frac{1}{\left(k-1\right)}\ne \frac{-1}{-\left(2k+1\right)}$
⇒ 3(k − 1) = 2k − 1 and −(2k + 1) ≠ −(k − 1)
⇒ 3k − 3 = 2k − 1 and −2k − 1 ≠ −k + 1
⇒ k = 2 and k ≠ −2
Thus, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ holds when k is equal to 2.
Hence, the given system of equations has no solution when the value of k is 2.