Question:
For what value of k, the following pair of linear equation has infinitely many solutions?
$10 x+5 y-(k-5)=0$
$20 x+10 y-k=0$
Solution:
The given equations are
$10 x+5 y-(k-5)=0$
$20 x+10 y-k=0$
$\frac{a_{1}}{a_{2}}=\frac{10}{20}, \frac{b_{1}}{b_{2}}=\frac{5}{10}, \frac{c_{1}}{c_{2}}=\frac{k-5}{k}$
For the equations to have infinite number of solutions
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Let us take
$\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\frac{5}{10}=\frac{k-5}{k}$
$5 \times k=10 \times(k-5)$
$50=10 k-5 k$
$50=5 k$
$\frac{50}{5}=k$
$10=k$
Hence, the value of $k=10$ when the pair of linear equations has infinitely many solutions.