The value of k for which the system of equations has no solution is
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

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.