2x + 3y = 7, (a + b) x + (2a – b) y = 3(a + b + 1)

Question

Find the values of a and b for which the following system of linear equations has infinitely many solutions: 

2x + 3y = 7, (a + b) x + (2a – b) y = 3(a + b + 1)

Comments

When Terms 1 and 3 are taken after obtaining the a-5b=0, how does Term 3 loose its negative value and become positive, if this was the case then why did term 2 not gain inverse polarity?

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In term 3 numerator is -7 and denominator is -3(a + b + 1).
Minus sign of both numerator and denominator got eliminated (as -1 x -1 =1) and third term becomes positive.

Answer 1

Given, pair of equations

2x + 3y = 7

and (a + b)x + (2a – b)y = 3(a + b + 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 = 2, b₁ = 3 and c₁ = – 7

and a_{2 =} (a + b), b₂ = (2a – b) and c₂ = – 3(a + b + 1)

For infinitely many solutions,

⇒ 2(2a – b) = 3(a + b)

⇒ 4a – 2b = 3a + 3b

⇒ 4a – 3a – 3b – 2b = 0

⇒ a – 5b = 0 …(1)

On taking I and III terms, we get

⇒ 6(a + b + 1) = 7(a + b)

⇒ 6a + 6b + 6 = 7a + 7b

⇒ 6a – 7a + 6b – 7b = – 6

⇒ – a – b = – 6

⇒ a + b = 6 …(2)

Solving eqⁿ (1) and (2), we get

⇒ b = 1

Now, substituting the value of b in eqⁿ (2), we get

⇒ a + b = 6

⇒ a + 1 = 6

⇒ a = 5