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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

Solution:

Given, the pair of linear equations are

x + 2y = 1

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

We have to determine the values of a and b for which the pair of linear equations has infinitely many solutions.

We know that,

For a pair of linear equations in two variables be a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\), then

i) The pair of linear equation is dependent and consistent

ii) The graph will be a pair of coincident lines. Each point on the lines will be a solution and so the pair of equations will have infinitely many solutions.

Here, a₁ = 1, b₁ = 2, c₁ = 1

a₂ = a - b, b₂ = a + b, c₂ = a + b - 2

So, a₁/a₂ = 1/(a - b)

b₁/b₂ = 2/(a + b)

c₁/c₂ = 1/(a+b-2)

For infinitely many solutions,

\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

1/(a - b) = 2/(a+b) = 1/(a + b - 2)

Case 1) 1/(a - b) = 2/(a + b)

1(a + b) = 2(a - b)

a + b = 2a - 2b

a - 2a + b + 2b = 0

-a + 3b = 0

a - 3b = 0

a = 3b -------------------------- (1)

Case 2) 2/(a + b) = 1/(a + b - 2)

2(a + b - 2) = a + b

2a + 2b - 4 = a + b

2a - a + 2b - b = 4

a + b = 4 --------------------------- (2)

Substitute (1) in (2),

3b + b = 4

4b = 4

b = 4/4

b = 1

Put b= 1 in (1),

a = 3(1)

a = 3

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

✦ Try This: For which value(s) of λ, do the pair of linear equations λx + y = λ and x + λy = 11 have infinitely many solutions

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 3

NCERT Exemplar Class 10 Maths Exercise 3.3 Problem 3

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

Summary:

For the values of a = 3 and b = 1, the pair of linear equations x + 2y = 1; (a - b)x + (a + b)y = a + b - 2 has infinitely many solutions.

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