Answer 1

2x + 3y -7 = 0

(a – b)x + (a + b)y - (3a +b –2) = 0

`a_1/a_2 = 2/(a-b) = 1/2`

`b_1/b_2 = 3/(a+b)`

`c_1/c_2 = -7/-(3a+b-2) = 7/(3a+b-2)`

For infinitely many solutions,

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

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

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

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

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

2a + 2b = 3a - 3b

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

Subtracting equation (i) from (ii), we get

4b = 4

b = 1

Putting this value in equation (ii), we get

a - 5 × 1 = 0

a = 5

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