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Find the value of k for which the system of linear equations has an infinite number of solutions:
2 x+(k-2) y=k
6 x+(2 k-1) y=(2 k+5)
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:
2 x+(k-2) y=k
6 x+(2 k-1) y=(2 k+5)

Solution:

The system of equations:
2 \mathrm{x}+(\mathrm{k}-2) \mathrm{y}=\mathrm{k}
\Rightarrow 2 \mathrm{x}+(\mathrm{k}-2) \mathrm{y}-\mathrm{k}=0\dots(i)
And, 6 \mathrm{x}+(2 \mathrm{k}-1) \mathrm{y}=(2 \mathrm{k}+5)
\Rightarrow 6 \mathrm{x}+(2 \mathrm{k}-1) \mathrm{y}-(2 \mathrm{k}+5)=0\dots(ii)
The 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}=2, b_{1}=(k-2), c_{1}=-k and a_{2}=6, b_{2}=(2 k-1), c_{2}=-(2 k+5)
For an infinite number of solutions, we must have:
\begin{array}{l} \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\ \frac{2}{6}=\frac{(k-2)}{(2 k-1)}=\frac{-k}{-(2 k+5)} \\ \Rightarrow \frac{1}{3}=\frac{(k-2)}{(2 k-1)}=\frac{k}{(2 k+5)} \end{array}
So now, we have the following three cases:

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

Case II:
\begin{array}{l} \frac{(k-2)}{(2 k-1)}=\frac{k}{(2 k+5)} \\ \Rightarrow(\mathrm{k}-2)(2 \mathrm{k}+5)=\mathrm{k}(2 \mathrm{k}-1) \\ \Rightarrow 2 \mathrm{k}^{2}+5 \mathrm{k}-4 \mathrm{k}-10=2 \mathrm{k}^{2}-\mathrm{k} \\ \Rightarrow \mathrm{k}+\mathrm{k}=10 \Rightarrow 2 \mathrm{k}=10 \Rightarrow \mathrm{k}=5 \end{array}

Case III:
\begin{array}{l} \frac{1}{3}=\frac{k}{(2 k+5)} \\ \Rightarrow 2 \mathrm{k}+5=3 \mathrm{k} \Rightarrow \mathrm{k}=5 \end{array}
As a result, the given system of equations has an infinite number of solutions when \mathrm{k} is equal to 5

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