Given, pair of equations
3x + 4y = 12
and (a + b)x + 2(a – b)y = 5a – 1
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 4 and c1 = – 12
and a2 = (a + b), b2 = 2(a – b) and c2 = – (5a – 1) = 1 + 5a
⇒ 6(a – b) = 4(a + b)
⇒ 6a – 6b = 4a + 4b
⇒ 6a – 4a – 6b – 4b = 0
⇒ 2a – 10b = 0
⇒ a – 5b = 0 …(1)
On taking I and III terms, we get
⇒ 3(5a – 1) = 12(a + b)
⇒ 15a – 3 = 12a + 12b
⇒ 15a – 12a – 12b = 3
⇒ 3a – 12b = 3
⇒ a – 4b = 1 …(2)
Solving eqn (1) and (2), we get
⇒ b = 1
Now, substituting the value of b in eqn (2), we get
⇒ a – 4b = 1
⇒ a – 4 = 1
⇒ a = 1 + 4
⇒ a = 5