For which values of a and b will the following pair of linear equations have infinitely many x+2y=1

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.

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]

`a_1/a_2 = 1/[a - b]`

`b_1/b_2 = 2/[a + b]`

`c_1/c_2 = 1/[a + b - 2]`

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

`a_1/a_2 = b_1/b_2 = c_1/c_2`   .....[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 equation [iii] in equation [iv], we get

3b + b = 4

4b = 4

b = 1

Put the value of b in equation [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.

`a=3, b=1``a=1, b=3``a=1, b=1``a=3 , b=3`

Answer : A

Solution : 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
`a_[1]=1, b_[1]=2` and `c_[1]=-1" "` [ from Eq. [i]]
`a_[2]=[a-b],b_[2]=[a+b] " " `[from Eq. [ii]]
and `" " c_[2] =-[a+b-2]`
For infinitely many solutions of the pairs of linear equations,
`[a_[1]]/[a_[2]]=[b_[1]]/[b_[2]]=[c_[1]]/[c_[2]]`
`rArr " " [1]/[a-b] =[2]/[a+b]=[-1]/[-[a+b-2]]`
Taking first two parts,
`[1]/[a-b]=[2]/[a+b]`
`rArr " " a+b = 2a-2b`
`rArr " " 2a-a=2b+b`
`rArr " " a = 3b " " ...[iii]`
Taking last two parts,
`[2]/[a+b]=[1]/[a+b-2]`
`rArr" " 2a+2b-4=a+b`
`rArr " " a+b=4 " " ...[iv]`
Now, put the value of a from Eq. [iii] in Eq. [iv], we get
`3b+b=4`
`rArr" "4b=4`
`rArr " " b = 1`
Put the value of b in Eq. [iii], we get
`a=3xx1`
`rArr " " 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.

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