Question

For which values of a and b does the following linear equations have infinite number

of solutions.

2x+3y = 7 and ( a-b)x + (a-b) y = 3a+b =2
​

Answer 1

The values of a and b for which the equations 2x + 3y = 7 and (a - b) x + (a + b) y = 3a + b - 2 will have infinitely many solutions will be a = 5 and b = 1.

Step-by-step explanation:

(i) 2x + 3y - 7 = 0

(a - b) x + (a + b) y - (3a + b - 2) = 0

a₁/a₂ = 2/(a - b)

b₁/b₂ = 3/(a + b)

c₁/c₂ = - 7/[-(3a + b - 2)] = 7/(3a + b - 2)

For infinitely many solutions,

a₁/a₂ = b₁/b₂ = c₁/c₂

2/(a - b) = 7/(3a + b - 2)

6a + 2b - 4 = 7a - 7b

a - 9b = - 4 ....(1)

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

2a + 2b = 3a - 3b

a - 5b = 0 ....(2)

Subtracting (1) from (2), we obtain

(a - 5b) - (a - 9b) = 0 - (-4)

4b = 4

b = 1

Substituting b = 1 in equation (2), we obtain

a - 5 × 1 = 0

a = 5

Hence, a = 5 and b = 1 are the values for which the given equations will have infinitely many solutions.