Question

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$$\begin{gathered} (k-1)x-y=5, \\ (k+1)x+(1-k)y=(3k+1). \end{gathered}$$

Answer 1

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

Now, we have the following three cases:
Case I:
$\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{1}{\left(k-1\right)}$
$\Rightarrow {\left(k-1\right)}^2=\left(k+1\right)$
$\Rightarrow k^2+1-2k=k+1$
$$\begin{gathered} \Rightarrow k^2-3k=0\Rightarrow k(k-3)=0 \\ \Rightarrow k=0\mathrm{or}k=3 \end{gathered}$$

Case II:
$\frac{1}{\left(k-1\right)}=\frac{5}{\left(3k+1\right)}$
$\Rightarrow 3k+1=5\left(k-1\right)$
$\Rightarrow 3k+1=5k-5$
$\Rightarrow 2k=6\Rightarrow k=3$

Case III:
$\frac{\left(k-1\right)}{\left(k+1\right)}=\frac{5}{\left(3k+1\right)}$
$\Rightarrow \left(3k+1\right)\left(k-1\right)=5\left(k+1\right)$
$\Rightarrow 3k^2+k-3k-1=5k+5$
$\Rightarrow 3k^2-2k-5k-1-5=0$
$\Rightarrow 3k^2-7k-6=0$
$\Rightarrow 3k^2-9k+2k-6=0$
$\Rightarrow 3k\left(k-3\right)+2\left(k-3\right)=0$
$\Rightarrow \left(k-3\right)\left(3k+2\right)=0$
$\Rightarrow \left(k-3\right)=0\mathrm{or}\left(3k+2\right)=0$
$\Rightarrow k=3\mathrm{or}\mathrm{k}=\frac{-2}{3}$

Hence, the given system of equations has an infinite number of solutions when k is equal to 3.