Question

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

Answer 1

Comparing $10x+5y-(k-5)=0$ with $a_{1}x+b_{1}y+c_1=0$,

and $20x+10y-k=0$ 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
$\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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

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

$\Rightarrow k=2k-10$

$\Rightarrow k=10$