Question

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$$\begin{gathered} x+(k+1)y=5, \\ (k+1)x+9y=(8k-1). \end{gathered}$$

Answer 1

The given system of equations:
x + (k + 1)y = 5
⇒ x + (k + 1)y − 5 = 0 ...(i)
And, (k + 1)x + 9y = (8k − 1)
⇒ (k + 1)x + 9y − (8k − 1) = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁= (k + 1), c₁ = −5 and a₂ = (k + 1), b₂ = 9, c₂ = −(8k − 1)
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{1}{\left(k+1\right)}=\frac{\left(k+1\right)}{9}=\frac{-5}{-\left(8k-1\right)}$
$\Rightarrow \frac{1}{\left(k+1\right)}=\frac{\left(k+1\right)}{9}=\frac{5}{\left(8k-1\right)}$

Now, we have the following three cases:
Case I:
$\frac{1}{\left(k+1\right)}=\frac{\left(k+1\right)}{9}$
$\Rightarrow {\left(k+1\right)}^2=9$
$\Rightarrow \left(k+1\right)=\pm 3$
$\Rightarrow \left(k+1\right)=3\mathrm{or}\left(k+1\right)=-3$
$\Rightarrow k=2\mathrm{or}\mathrm{k}=-4$

Case II:
$\frac{k+1}{9}=\frac{5}{8k-1}$
$\Rightarrow \left(k+1\right)\left(8k-1\right)=45$
$\Rightarrow 8k^2-k+8k-1=45$
$\Rightarrow 8k^2+7k-46=0$
$\Rightarrow 8k^2+23k-16k-46=0$
$\Rightarrow k\left(8k+23\right)-2\left(8k+23\right)=0$
$\Rightarrow \left(k-2\right)\left(8k+23\right)=0$
$\Rightarrow \left(k-2\right)=0\mathrm{or}\left(8k+23\right)=0$
$\Rightarrow k=2\mathrm{or}k=\frac{-23}{8}$

Case III:
$\frac{1}{\left(k+1\right)}=\frac{5}{\left(8k-1\right)}$
$\Rightarrow \left(8k-1\right)=5\left(k+1\right)$
$\Rightarrow \left(8k-1\right)=5k+5$
$\Rightarrow 3k=6\Rightarrow k=2$

Hence, the given system of equations has an infinite number of solutions when k is equal to 2.