Question:
The value of k for which the system of equations has no solution is
$x+2 y=5$
$3 x+k y+15=0$
(a) 6
(b) $-6$
(c) $3 / 2$
(d) None of these
Solution:
The given system of equation is
$x+2 y=5$
$3 x+k y+15=0$
If $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ then the equation have no solution.
$\frac{1}{3}=\frac{2}{k}=\frac{-5}{15}$
By cross multiply we get
$k \times 1=3 \times 2$
$k=6$
Hence, the correct choice is a