For which values of A and B will the following pair of linear equations has infinitely many solutions x 2y 1 and a B x a by a/b 2?

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.

Solution

Step 1: Convert the equations to standard form

Given two lines are,

x+2y=1 and (a-b)x+(a+b)y=a+b-2

The standard linear equation is given as,

ax+by+c=0

Then, the two equations in standard form are given as,

x+2y=1⇒x+2y- 1=0

(a-b)x+(a+b)y=a+b-2⇒(a-b)x+(a+b)y-a -b+2=0

Step 2: Compare the equations with the standard equation

Comparing the two equations with the standard equation we define,

a1=1, b1=2 and c1=-1

a2=a-b, b2=a+b and c2=-(a+b-2 )

Step 3: Defining the condition for infinite solutions

For infinitely many solutions of the pair of linear equations condition is satisfied by,

a1a2=b1b2=c1c2

Thus,

1a-b=2a+b=1a+b-2

We have,

1a-b=2a+b⇒2(a-b)=a+b⇒ a=2b

We also have,

2a+b=1a+b-2 ⇒a+b=2(a+b-2)⇒a+b=4

Substituting the above equation in the previous equation,

2b+b=4 ⇒b=1

Thus, a=2×1=2

Therefore, when a=2 and b=1, the given set 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.

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.

For which values of A and B will the following pair of linear equation has infinitely many solutions?

Therefore, for the values of a = 3 and b = 1, the pair of linear equations has infinitely many solutions.

For which values of A and B will the following pair of linear equations has infinitely many solutions x 2y 1 and a B x a b'y a/b 2?

Hence, required values of a and b are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.

For which values of A and B will the following pair of linear equations have infinitely many 2x 3y 7?

Summary: (i) The values of a and b for which the equations 2x + 3y = 7 and (a - b) x + (a + b) y = 3a + b - 2 will have infinitely many solutions will be a = 5 and b = 1.

How do you know if a linear equation has infinite solutions?

Conditions for Infinite Solution If the two lines have the same y-intercept and the slope, they are actually in the same exact line. In other words, when the two lines are the same line, then the system should have infinite solutions.

Chủ đề