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.
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