# EXAMPLES OF SOLVING A PAIR OF LINEAR EQUATIONS WITH GIVEN CONDITION

If a pair of linear equations is having the types of solutions given below, then it has to meet out the corresponding conditions.

 Types of Solutionunique solutioninfinitely many solutions no solution Condition a1/a2  ≠  b1/b2 a1/a2  =  a1/a2  = c1/c2a1/a2  =  a1/a2  ≠  c1/c2

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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