Find the value of $k$ for which the following system of linear equations has infinite solutions
$k x + 3 y = 2 k + 1 : 2 (k + 1) x + 9 y = 7 k + 1$

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Solution

Step 1:If the system of linear equations has infinite solutions
Applying conditions
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
Given,
$k x + 3 y = 2 k + 1 \Rightarrow k x + 3 y - (2 k + 1) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) 2 (k + 1) x + 9 y = 7 k + 1 \Rightarrow 2 (k + 1) x + 9 y - (7 k + 1) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$
Equation (1) and (2) are in standard forms
$h e r e a_{1} = k b_{1} = 3 c_{1} = - (2 k + 1) a_{2} = 2 (k + 1) b_{2} = 9 c_{2} = - (7 k + 1) i . e \frac{k}{2 (k + 1)} = \frac{1}{3} = \frac{2 k + 1}{7 k + 1}$
Step 2:To find the value of $k$ using cross multiplication method
$T a k i n g \frac{k}{2 (k + 1)} = \frac{1}{3} \Rightarrow 3 k = 2 k + 2 \Rightarrow 3 k - 2 k = 2 \Rightarrow k = 2$
Step 3:Verification
$\frac{k}{2 (k + 1)} = \frac{2}{2 (2 + 1)} = \frac{1}{3} \frac{2 k + 1}{7 k + 1} = \frac{5}{15} = \frac{1}{3} t h e r e f o r e, \frac{k}{2 (k + 1)} = \frac{1}{3} = \frac{2 k + 1}{7 k + 1}$
Therefore the value of $k = 2$

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