**Question:**

Write the value of $k$ for which the system of equations $x+y-4=0$ and $2 x+k y-3=0$ has no solution.

**Solution:**

The given system of equations is

$x+y-4=0$

$2 x+k y-3=0$

$a_{1}=1, a_{2}=2, b_{1}=1, b_{2}=k, c_{1}=4, c_{2}=3$

For the equations to have no solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\frac{1}{2}=\frac{1}{k}$

By cross multiplication we get,

$1 \times k=1 \times 2$

$k=2$

Hence, the value of $k$ is 2 when system equations has no solution.