Question -
Answer -
Solution:
The given pair oflinear 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 manysolutions 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 twoparts,
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 ofa from Eq. (iii) in Eq. (iv), we get
3b + b = 4
4b = 4
b = 1
Put the value of b inEq. (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 1respectively for which the given pair of linear equations has infinitely manysolutions.