Question

If 2x – 3y = 7 and (a + b)x – (a + b – 3)y = 4a + b have an infinite number of solutions then

Options

• a = 5, b = 1

• a = –5, b = 1

• a = 5, b = –1

• a = –5, b = –1

Answer 1

a = –5, b = –1

Explanation:

1. Identify system coefficients

For a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, an infinite number of solutions exists if and only if their corresponding ratios are equal:

`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`

Rewriting the given equations into standard form:

Equation 1: 2x – 3y – 7 = 0

⇒ a₁ = 2, b₁ = –3, c₁ = –7

Equation 2: (a + b)x – (a + b – 3)y – (4a + b) = 0

⇒ a₂ = a + b, b₂ = –(a + b – 3), c₂ = –(4a + b)

Substituting these values into the ratio condition gives:

`2/(a + b) = (-3)/(-(a + b - 3))`

= `(-7)/(-(4a + b))`

⇒ `2/(a + b) = (-3)/(a + b - 3)`

= `(7)/(4a + b)`

2. Set up equations

Equating the first two ratios to find our first relation:

`2/(a + b) = 3/(a + b - 3)`

Cross-multiplying:

2(a + b – 3) = 3(a + b)

2a + 2b – 6 = 3a + 3b

a + b = –6   ...(Equation I)

Equating the first and third ratios to find our second relation:

`2/(a + b) = 7/(4a + b)`

Cross-multiplying:

2(4a + b) = 7(a + b)

8a + 2b = 7a + 7b

a = 5b   ...(Equation II)

3. Solve for variables

Substitute Equation II into Equation I:

5b + b = –6

6b = –6

⇒ b = –1

Now substitute b = –1 back into Equation II:

a = 5(–1)

⇒ a = –5