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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