Find the value of $k$ for which the given system of equations has infinite number of solutions. $10 x + 5 y - (k - 5) = 0$ and $20 x + 10 y - k = 0$

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Solution

Comparing $10 x + 5 y - (k - 5) = 0$ with $a_{1} x + b_{1} y + c_{1} = 0$ ,

and $20 x + 10 y - k = 0$ with $a_{2} x + b_{2} y + c_{2} = 0$ ,

Here, $a_{1} = 10, b_{1} = 5, c_{1} = - (k - 5)$
$a_{2} = 20, b_{2} = 10, c_{2} = - k$
Given the system has infinitely many solution
$∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{10}{20} = \frac{5}{10} = \frac{k - 5}{k}$
$\Rightarrow \frac{1}{2} = \frac{k - 5}{k}$
$\Rightarrow k = 2 k - 10$
$\Rightarrow k = 10$

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