Question

If 2a = b, the pair of equations ax + by = 2a² – 3b² , x + 2y = 2a – 6b possess : 

(a) no solution 

(b) only one solution 

(c) only two solutions 

(d) an infinite number of solutions

Answer 1

(d) an infinite number of solutions

Here a₁ = a, b₁ = b, c₁ = – (2a² – 3b² ) = 3b² – 2a²

a₂ = 1, b₂ = 2, c₂ = – (2a – 6b) = 6b – 2a

\(\frac{a_1}{a_2}=\frac{a}{1}\), \(\frac{b_1}{b_2}=\frac{b}{2}\),\(\frac{c_1}{c_2}\)= \(\frac{3b^2-2a^2}{6b-2a}\)

2a = b \(\Rightarrow\) a = \(\frac{b}{2}\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac{b}{2}\), \(\frac{b_1}{b_2}=\frac{b}{2}\) and 

\(\frac{c_1}{c_2}\)= \(\frac{3b^2-2\times\frac{b^2}{4}}{6b-2\times\frac{b}{2}}\)= \(\frac{3b^2-\frac{b^2}{2}}{6b-b}\) = \(\frac{\frac{5b^2}{2}}{5b}\)=\(\frac{b}{2}\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)