Question:
The value of $k$ for which the system of equations $x+2 y-3=0$ and $5 x+k y+7=0$ has no solution, is
(a) 10
(b) 6
(c) 3
(d) 1
Solution:
The given system of equations are
$x+2 y-3=0$
$5 x+k y+7=0$
For the equations to have no solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\frac{1}{5}=\frac{2}{k}=\frac{-3}{7}$
If we take
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$
$\frac{1}{5}=\frac{2}{k}$
$k=10$
Therefore the value of k is10.
Hence, correct choice is $a$.