Question

The system $x+2y=3\mathrm{and}5x+ky+7=0$ has no solution, when

(a) k = 10
(b) $k\ne 10$
(c) $k=\frac{-7}{3}$
(d) k = −21

Answer 1

The correct option is (a).

The given system of equations can be written as follows:
x + 2y − 3 = 0 and 5x + ky + 7 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁ = 2, c₁ = −3 and a₂ = 5, b₂ = k and c₂ = 7
∴ $\frac{a_1}{a_2}=\frac{1}{5},\frac{b_1}{b_2}=\frac{2}{k}\mathrm{and}\frac{c_1}{c_2}=\frac{-3}{7}$
For the system of equations to have no solution, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
∴ $\frac{1}{5}=\frac{2}{k}\ne \frac{-3}{7}\Rightarrow k=10$