Question

3. For which values of a and b, will the following pair of linear equations have infinitely many
solutions?
x + 2y = 1
(a – b)x + (a + b)y = a + b – 2

Answer 1

Answer:

The given pair of linear equations are:

x + 2y = 1 …(i)

(a-b)x + (a + b)y = a + b - 2 …(ii)

On comparing with ax + by = c = 0 we get

a1 = 1, b1 = 2, c1 = - 1

a2 = (a - b), b2 = (a + b), c2 = - (a + b - 2)

a1 /a2 = 1/(a-b)

b1 /b2 = 2/(a+b)

c1 /c2 = 1/(a+b-2)

For infinitely many solutions of the, pair of linear equations,

a1/a2 = b1/b2=c1/c2(coincident lines)

so, 1/(a-b) = 2/ (a+b) = 1/(a+b-2)

Taking first two parts,

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

a + b = 2(a - b)

a = 3b …(iii)

Taking last two parts,

2/ (a+b) = 1/(a+b-2)

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

a + b = 4 …(iv)

Now, put the value of a from Eq. (iii) in Eq. (iv), we get

3b + b = 4

4b = 4

b = 1

Put the value of b in Eq. (iii), we get

a = 3

So, the values (a,b) = (3,1) satisfies all the parts. Hence, required values of a and b are 3 and 1

respectively for which the given pair of linear equations has infinitely many solutions