Question

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:

Given pair of linear equations are

x + 2y = 1   ...(i)

and (a−b)x+(a+b)y=a+b−2 ...(ii)

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

a1=1,b1=2 and c1=−1  [ from Eq. (i)]

a2=(a−b),b2=(a+b) [from Eq. (ii)]

and  c2=−(a+b−2)

For infinitely many solutions of the pairs of linear equations,

a1a2=b1b2=c1c2

⇒ 1a−b=2a+b=−1−(a+b−2)

Taking first two parts,

1a−b=2a+b

⇒ a+b=2a−2b

⇒ 2a−a=2b+b

⇒ a=3b ...(iii)

Taking last two parts,

2a+b=1a+b−2

⇒ 2a+2b−4=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=3×1

⇒ a=3

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

Step-by-step explanation:

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