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Find the value of k for which the system of linear equations has an infinite number of solutions:
5 x+2 y=2 k
2(k+1) x+k y=(3 k+4)
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:
5 x+2 y=2 k
2(k+1) x+k y=(3 k+4)

Solution:

Given system of equations:
\begin{array}{l} 5 x+2 y=2 k \\ \Rightarrow 5 x+2 y-2 k=0\dots \dots(i) \end{array}
And, 2(\mathrm{k}+1) \mathrm{x}+\mathrm{ky}=(3 \mathrm{k}+4)
\Rightarrow 2(\mathrm{k}+1) \mathrm{x}+\mathrm{ky}-(3 \mathrm{k}+4)=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}=5, b_{1}=2, c_{1}=-2 k and a_{2}=2(k+1), b_{2}=k, c_{2}=-(3 k+4)
For an infinite no. of solutions, we must have:
\begin{array}{l} \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\ \frac{5}{2(k+1)}=\frac{2}{k}=\frac{-2 k}{-(3 k+4)} \\ \Rightarrow \frac{5}{2(k+1)}=\frac{2}{k}=\frac{2 k}{(3 k+4)} \end{array}
We have the following three cases:

Case I:
\begin{array}{l} \frac{5}{2(k+1)}=\frac{2}{k} \\ \Rightarrow 2 \times 2(k+1)=5 k \\ \Rightarrow 4(k+1)=5 k \\ \Rightarrow 4 k+4=5 k \\ \Rightarrow k=4 \end{array}

Case II:
\frac{2}{k}=\frac{2 k}{(3 k+4)}
\Rightarrow 2 \mathrm{k}^{2}=2 \times(3 \mathrm{k}+4)
\Rightarrow 2 \mathrm{k}^{2}=6 \mathrm{k}+8 \Rightarrow 2 \mathrm{k}^{2}-6 \mathrm{k}-8=0
\Rightarrow 2\left(\mathrm{k}^{2}-3 \mathrm{k}-4\right)=0
\Rightarrow \mathrm{k}^{2}-4 \mathrm{k}+\mathrm{k}-4=0
\Rightarrow \mathrm{k}(\mathrm{k}-4)+1(\mathrm{k}-4)=0
\Rightarrow(\mathrm{k}+1)(\mathrm{k}-4)=0
\Rightarrow(\mathrm{k}+1)=0 or (\mathrm{k}-4)=0
\Rightarrow \mathrm{k}=-1 \text { or } \mathrm{k}=4

Case III:
\begin{array}{l} \frac{5}{2(k+1)}=\frac{2 k}{(3 k+4)} \\ \Rightarrow 15 \mathrm{k}+20=4 \mathrm{k}^{2}+4 \mathrm{k} \\ \Rightarrow 4 \mathrm{k}^{2}-11 \mathrm{k}-20=0 \\ \Rightarrow 4 \mathrm{k}^{2}-16 \mathrm{k}+5 \mathrm{k}-20=0 \\ \Rightarrow 4 \mathrm{k}(\mathrm{k}-4)+5(\mathrm{k}-4)=0 \\ \Rightarrow(\mathrm{k}-4)(4 \mathrm{k}+5)=0 \\ \Rightarrow \mathrm{k}=4 \text { or } \mathrm{k}=\frac{-5}{4} \end{array}
As a result, the given system of equations has an infinite no. of solutions when \mathrm{k} is equal to 4.

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