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Find the value of k for which the system of linear equations has an infinite number of solutions:
(\mathrm{k}-1) \mathrm{x}-\mathrm{y}=5
(\mathrm{k}+1) \mathrm{x}+(1-\mathrm{k}) \mathrm{y}=(3 \mathrm{k}+1)
CBSE | Class 10 | Excercise 3D | Maths | Pair of Linear Equations in Two Variables | RS Aggarwal
Find the value of k for which the system of linear equations has an infinite number of solutions:
(\mathrm{k}-1) \mathrm{x}-\mathrm{y}=5
(\mathrm{k}+1) \mathrm{x}+(1-\mathrm{k}) \mathrm{y}=(3 \mathrm{k}+1)

Solution:

Given system of equations:
\begin{array}{l} (\mathrm{k}-1) \mathrm{x}-\mathrm{y}=5 \\ \Rightarrow(\mathrm{k}-1) \mathrm{x}-\mathrm{y}-5=0\dots \dots(i) \end{array}
And, (k+1) x+(1-k) y=(3 k+1)
\Rightarrow(\mathrm{k}+1) \mathrm{x}+(1-\mathrm{k}) \mathrm{y}-(3 \mathrm{k}+1)=0\dots \dots(ii)
Equations are of the following form:
a_{1} x+b_{1} y+c_{1}=0, a_{2} x+b_{2} y+c_{2}=0
where, a_{1}=(k-1), b_{1}=-1, c_{1}=-5 and a_{2}=(k+1), b_{2}=(1-k), c_{2}=-(3 k+1)
For an infinite no. of solutions, we must have:
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}
that is, \frac{(\mathrm{k}-1)}{(k+1)}=\frac{-1}{-(k-1)}=\frac{-5}{-(3 k+1)}
\Rightarrow \frac{(\mathrm{k}-1)}{(k+1)}=\frac{1}{(k-1)}=\frac{5}{(3 k+1)}
We have the following three cases:

Case I:
\begin{array}{l} \frac{(k-1)}{(k+1)}=\frac{1}{(k-1)} \\ \Rightarrow(k-1)^{2}=(k+1) \\ \Rightarrow k^{2}+1-2 k=k+1 \end{array}
\begin{array}{l} \Rightarrow \mathrm{k}^{2}-3 \mathrm{k}=0 \Rightarrow \mathrm{k}(\mathrm{k}-3)=0 \\ \Rightarrow \mathrm{k}=0 \text { or } \mathrm{k}=3 \end{array}

Case II:
\begin{array}{l} \frac{1}{(k-1)}=\frac{5}{(3 k+1)} \\ \Rightarrow 3 \mathrm{k}+1=5 \mathrm{k}-5 \\ \Rightarrow 2 \mathrm{k}=6 \Rightarrow \mathrm{k}=3 \end{array}

Case III:
\begin{array}{l} \frac{(\mathrm{k}-1)}{(k+1)}=\frac{5}{(3 k+1)} \\ \Rightarrow(3 \mathrm{k}+1)(\mathrm{k}-1)=5(\mathrm{k}+1) \\ \Rightarrow 3 \mathrm{k}^{2}+\mathrm{k}-3 \mathrm{k}-1=5 \mathrm{k}+5 \\ \Rightarrow 3 \mathrm{k}^{2}-2 \mathrm{k}-5 \mathrm{k}-1-5=0 \\ \Rightarrow 3 \mathrm{k}^{2}-7 \mathrm{k}-6=0 \\ \Rightarrow 3 \mathrm{k}^{2}-9 \mathrm{k}+2 \mathrm{k}-6=0 \\ \Rightarrow 3 \mathrm{k}(\mathrm{k}-3)+2(\mathrm{k}-3)=0 \\ \Rightarrow(\mathrm{k}-3)(3 \mathrm{k}+2)=0 \\ \Rightarrow(\mathrm{k}-3)=0 \text { or }(3 \mathrm{k}+2)=0 \\ \Rightarrow \mathrm{k}=3 \text { or } \mathrm{k}=\frac{-2}{3} \end{array}
As a result, the given system of equations has an infinite no. of solutions when \mathrm{k} is equal to 3 .

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