Find the value of $c$ for which the following system of linear equations has infinite solution
$2 x + (c - 2) y = c; 6 x + (2 c - 1) y = 2 c + 5$

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Solution

Step 1:If the system of linear equations has infinite solutions
Applying condition
$\frac{a_{1}}{a 2} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
Given,
$2 x + (c - 2) y = c \Rightarrow 2 x + (c - 2) y - c = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) 6 x + (2 c - 1) y = 2 c + 5 \Rightarrow 6 x + (2 c - 1) y - (2 c + 5) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$
Equation (1) and (2) are in standard form
$h e r e a_{1} = 2 b_{1} = c - 2 c_{1} = - c a_{2} = 6 b_{2} = 2 c - 1 c_{2} = - (2 c + 5) i . e \frac{2}{6} = \frac{c - 2}{2 c - 1} = \frac{- c}{- (2 c + 5)} \Rightarrow \frac{1}{3} = \frac{c - 2}{2 c - 1} = \frac{c}{2 c + 5}$
Step 2:To find the value of $c$ by cross multiplication method
$T a k i n g \frac{1}{3} = \frac{c - 2}{2 c - 1} \Rightarrow 2 c - 1 = 3 (c - 2) \Rightarrow 2 c - 1 = 3 c - 6 \Rightarrow 3 c - 2 c = 6 - 1 \Rightarrow c = 5$
Step 3:Verification
$\frac{c - 2}{2 c - 1} = \frac{5 - 2}{10 - 1} = \frac{3}{9} = \frac{1}{3} \frac{c}{2 c + 5} = \frac{5}{15} = \frac{1}{3} t h e r e f o r e, \frac{1}{3} = \frac{c - 2}{2 c - 1} = \frac{c}{2 c + 5}$
Therefore the value of $c = 5$

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