Question

For what value of k does the following pair of linear equations have infinitely many solutions?:
$10x+5y-\left(k-5\right)=0,20x+10y-k=0$

Answer 1

The given pair of linear equations are
$$\begin{gathered} 10x+5y-\left(k-5\right)=0.....\left(i\right) \\ 20x+10y-k=0.....\left(ii\right) \end{gathered}$$
Which is of the form $a_{1}x+b_{1}y+c_1=0\mathrm{and}{a}_{2}x+b_{2}y+c_2=0$, where $a_1=10,b_1=5,c_1=-\left(k-5\right),a_2=20,b_2=10\mathrm{and}{c}_2=-k$
For the given pair of linear equations to have infinitely many solutions, we must have
$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \Rightarrow \frac{10}{20}=\frac{5}{10}=\frac{-\left(k-5\right)}{-k} \\ \Rightarrow \frac{1}{2}=\frac{k-5}{k} \\ \Rightarrow 2k-10=k\Rightarrow k=10 \end{gathered}$$
Hence, k = 10.