**Examples of Solving a Pair of Linear Equations with Given Condition :**

In this section, we will learn how to solve 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.

unique solution infinitely many solutions no solution |
a a a |

**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 = – (3 a + 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

After having gone through the stuff and examples, we hope that the students would have understood, solving a pair of linear equations with given condition

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