Question

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$$\begin{gathered} 2x+3y=7, \\ (k-1)x+(k+2)y=3k. \end{gathered}$$

Answer 1

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0 ....(i)
And, (k − 1)x + (k + 2)y = 3k
⇒ (k − 1)x + (k + 2)y − 3k = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁= 3, c₁ = −7 and a₂ = (k − 1), b₂ = (k + 2), c₂ = −3k
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{2}{\left(k-1\right)}=\frac{3}{\left(k+2\right)}=\frac{-7}{-3k}$
$\Rightarrow \frac{2}{\left(k-1\right)}=\frac{3}{\left(k+2\right)}=\frac{7}{3k}$

Now, we have the following three cases:
Case I:
$\frac{2}{k-1}=\frac{3}{k+2}$
$\Rightarrow 2\left(k+2\right)=3\left(k-1\right)\Rightarrow 2k+4=3k-3\Rightarrow k=7$

Case II:
$\frac{3}{k+2}=\frac{7}{3k}$
$\Rightarrow 7\left(k+2\right)=9k\Rightarrow 7k+14=9k\Rightarrow 2k=14\Rightarrow k=7$

Case III:
$\frac{2}{k-1}=\frac{7}{3k}$
$\Rightarrow 7k-7=6k\Rightarrow k=7$

Hence, the given system of equations has an infinite number of solutions when k is equal to 7.