Question

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$$\begin{gathered} 2x+(k-2)y=k, \\ 6+(2k-1)y=(2k+5). \end{gathered}$$

Answer 1

The given system of equations:
2x + (k − 2)y = k
⇒ 2x + (k − 2)y − k = 0 ....(i)
And, 6x + (2k − 1)y = (2k + 5)
⇒ 6x + (2k − 1)y − (2k + 5) = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁= (k − 2), c₁ = −k and a₂ = 6, b₂ = (2k − 1), c₂ = −(2k + 5)
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{2}{6}=\frac{\left(k-2\right)}{\left(2k-1\right)}=\frac{-k}{-\left(2k+5\right)}$
$\Rightarrow \frac{1}{3}=\frac{\left(k-2\right)}{\left(2k-1\right)}=\frac{k}{\left(2k+5\right)}$

Now, we have the following three cases:
Case I:
$\frac{1}{3}=\frac{k-2}{2k-1}$
$\Rightarrow (2k-1)=3(k-2)$
$\Rightarrow 2k-1=3k-6\Rightarrow k=5$

Case II:
$\frac{k-2}{2k-1}=\frac{k}{2k+5}$
$\Rightarrow \left(k-2\right)\left(2k+5\right)=k\left(2k-1\right)$
$\Rightarrow 2k^2+5k-4k-10=2k^2-k$
$\Rightarrow k+k=10\Rightarrow 2k=10\Rightarrow k=5$

Case III:
$\frac{1}{3}=\frac{k}{2k+5}$
$\Rightarrow 2k+5=3k\Rightarrow k=5$

Hence, the given system of equations has an infinite number of solutions when k is equal to 5.