Question

The pair of linear equations 4x − 5y −20 = 0 and 3x + 5y − 15 = 0 has
(a) a unique solution
(b) two solutions
(c) many solutions
(d) no solution

Answer 1

(a) a unique solution
The given equations are of the form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where
a₁ = 4, b₁ = - 5, c₁ = - 20 and a₂ = 3, b₂ = 5 and c₂ = - 15
∴ $\frac{a_1}{a_2}=\frac{4}{3}$, $\frac{b_1}{b_2}=\frac{-5}{5}$ and $\frac{c_1}{c_2}=\frac{-20}{-15}=\frac{4}{3}$
∴ These graph lines will intersect at a unique point, when $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$.
Hence, the given system has a unique solution.