Question

For which values of a and b, will the following pair of linear equations have infinitey many
solutions?
x+2y=1
(a−b)x+ (a−b)y=a+b−2​

Answer 1

Step-by-step explanation:

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