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.