Question

Find the values of a and b for which the following system of equations has infinitely
many solutions.
(2a-1)x-3y = 5
3x+(b-2)y = 3

Answer 1

❥︎❥︎ANSWER࿐

Given, pair of equations

(2a – 1)x – 3y = 5

and 3x + (b – 2)y = 3

On comparing the given equation with standard form i.e.

a1 x + b1y + c1 = 0

and a2 x + b2y + c2 = 0,q we get

a1 = (2a – 1), b1 = – 3 and c1 = – 5

and a2 = 3, b2 = b – 2 and c2 = 3

For infinitely many solutions,

⇒ 3(2a – 1) = 15

⇒ 6a – 3 = 15

⇒ 6a = 15 +3

⇒ a = 18/6 = 3

On taking II and III terms, we get

$\frac{2a - 1}{3} = \frac{5}{3}$

⇒ -3/b - 2 = 5/3

⇒ – 9 = 5(b – 2)

⇒ 5b – 10 = – 9

⇒ 5b = – 9 + 10

⇒ b = 1/5.

Answer 2

❀❀ANSWER❀❀

Given, pair of equations

(2a – 1)x – 3y = 5 and 3x + (b – 2)y = 3

On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0

and a2 x + b2y + c2 = 0, we get

a1 = (2a – 1), b1 = – 3 and c1 = – 5

and a2 = 3, b2 = b – 2 and c2 = 3

$\frac{2a - 1}{3} = \frac{5}{3}$

⇒ 3(2a – 1) = 15

⇒ 6a – 3 = 15

⇒ 6a = 15 + 3

⇒ a = 18/6 = 3

On taking II and III terms, we get

⇒ -3/b - 2 = 5/3

⇒ – 9 = 5(b – 2)

⇒ 5b – 10 = – 9

⇒ 5b = – 9 + 10

⇒ b = 1/5