Question

For what value of k, the system of equations has no solution?
$x+2y=3$
$3x+ky-15=0$

• A.
• B.
• C.
• D. No values of k

Answer 1

The correct option is C

Condition for no solution: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

For the given lines

$x+2y=3$
$3x+ky-15=0$

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

When $k=6$ the set of equations represent two parallel lines.and therefore has no solution.