Question

Find the value of k for which the given system of equations has no solution?
$x+2y=3$
$3x+ky-15=0$

• A. $k=6$
• B. None of the above
• C. $k\ne 6$
• D. $k\ne 7$

Answer 1

The correct option is A $k=6$

Give: $x+2y=3$
$3x+ky-15=0$
Condition for no solution:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

$a_1=1,a_2=3,b_1=2,b_2=k$

$\frac{1}{3}=\frac{2}{k}\ne \frac{3}{15}\Rightarrow k=6$

When $k=6$ the set of equations represents two parallel lines which don't intersect at any point. So they can't have a common point which would be the solution.