EXAMPLES OF SOLVING A PAIR OF LINEAR EQUATIONS WITH GIVEN CONDITION

Example 1 :

For which values of a and b does the following pair of linear equations have an infinite many solution

2x + 3y = 7

(a – b) x + (a + b) y = 3 a + b – 2

Solution :

Condition for having infinitely many solutions

a₁/a₂  =  b₁/b₂  =  c₁/c₂

2 x + 3 y – 7 = 0 --------(1)

(a – b) x + (a + b) y – (3 a + b – 2) = 0 --------(2)

From the above information let us take the values of a₁ , a₂, b₁, b₂, c₁ and c ₂

 a₁  =  2               b₁  =  3                      c₁  =  -7

 a₂  =  (a-b)          b₂  =  (a + b)             c₂  =  – (3a + b – 2)

2/(a – b)  =  3/(a + b)  =  -7/-(3a + b – 2)

 2/(a – b)  =  3/(a + b)  =  7/(3a + b – 2)

2/(a – b)  =  7/(3a + b – 2)

2(3a + b – 2)  =  7(a – b)

6 a + 2 b – 4  =  7 a – 7 b

6 a – 7 a + 2 b + 7 b  =  4

-a + 9 b  =  4  ----- (3)

3(3a + b – 2)  =  7(a + b)

9 a + 3 b – 6  =  7 a + 7 b

9 a – 7 a + 3 b – 7 b  =  6

2 a – 4 b  =  6 ----- (4)

(3) ⋅ 2 + (4)

-2a + 18b  =  8

2a - 4b  =  6

---------------

14b  =  14

b  =  1

By applying the value of b in (3), we get

-a + 9(1)  =  4

-a + 9  =  4

-a  =  4 - 9

a  =  5

Example 2 :

For which value of k will the following pair of linear equations have no solution

3 x + y = 1

(2k – 1) x + (k – 1) y = 2 k + 1

Solution :

3x + y – 1  =  0  --------(1)

(2k – 1)x + (k – 1)y – (2 k + 1) = 0  --------(2)

Condition for having no solution                           

a₁/a ₂ = b₁/b ₂ ≠ c₁/c ₂

From the above information let us take the values of a₁ , a₂, b₁, b₂, c₁ and c ₂

 a₁ = 3                    b₁ = 1                    c₁ = -1

 a₂ = (2 k - 1)         b₂ = (k - 1)              c₂ = – (2 k + 1)

3/(2 k – 1)  =  1/(k – 1)

3 (k  - 1)  =  1 (2 k- 1)

3 k – 3  =  2 k – 1

3 k – 2 k  =  -1 + 3

k = 2

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