Question:
Write the value of $k$ for which the system of equations $3 x-2 y=0$ and $k x+5 y=0$ has infinitely may solutions.
Solution:
The given equations are
$3 x-2 y=0$
$k x+5 y=0$
$\frac{a_{1}}{a_{2}}=\frac{3}{k}, \frac{b_{1}}{b_{2}}=\frac{-2}{5}, \frac{c_{1}}{c_{2}}=\frac{0}{0}$
For the equations to have infinite number of solutions, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Therefore,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$
$\frac{3}{k}=\frac{-2}{5}$
By cross multiplication we have
$3 \times 5=-2 \times k$
$15=-2 k$
$\frac{15}{-2}=k$
Hence, the value of $k$ for the system of equation $3 \times-2 y=0$ and $k x+5 y=0$ is $\frac{-15}{2}$.