# TYPES OF SOLUTIONS OF PAIR OF LINEAR EQUATION IN TWO VARIABLES

Types of Solutions of Pair of Linear Equations in Two Variables :

In this section, we will learn the types of solution of a pair of linear equations are having. By comparing the coefficients of linear equations, we may find what type of solution they have.

## Types of Solutions of Pair of Linear Equations in Two Variables

To compare the coefficients of linear equations in two variables, the equations must be in the form.

a1x + b1y + c1  =  0

a2x + b2y + c2  =  0

The following three cases are possible for any given system of linear equations.

(i)  a1/a2    b1/b2, we get a unique solution

(ii)  a1/a2  =  a1/a = c1/c2, there are infinitely many solutions.

(iii)  a1/a2  =  a1/a ≠  c1/c2, there is no solution

## Verify the Condition for Consistency of a System of Linear Equations in Two Variables - Examples

Example 1 :

Which of the following pairs of linear equations has unique solution, no solution, infinitely many solutions. In case there is unique solution, find it by using cross multiplication method.

(i)   x – 3y – 3  =  0

3x – 9y – 2  =  0

Solution :

From the above information, let us take the values of a1, a2, b1, b2, c1 and c2

a1  =  1, b1  =  -3, c1  =  -3

a1  =  3, b1  =  -9, c1  =  -2

a1/a2  =  1/3

b1/b =  -3/(-9)  =  1/3

c1/c =  -3/(-2)  =  3/2

Here a1/a =  b1/b ≠  c1/c2

From this we can decide the given lines are parallel.

(ii)  2x + y  =  5

3x + 2y  =  8

Solution :

2x + y – 5  =  0

3x + 2y – 8  =  0

From the above information, let us take the values of a1, a2, b1, b2, c1 and c2

a1  =  2, b1  =  1, c1  =  -5

a1  =  3, b1  =  2, c1  =  -8

a1/a =  2/3

b1/b =  1/2

c1/c2  =  (-5)/(-8) = 5/8

Here, a₁/a₂  ≠  b₁/b₂

Therefore two given lines are intersecting

x/(-8  + 10)  =  y/(-15 + 16)  =  1/(4 – 3)

x/2  =  y/1  =  1/1

x/2  =  1           y/1  =  1

x  =  2           y  =  1

(iii) 3x – 5y  =  20

6x – 10y  =  40

Solution :

3 x – 5 y – 20  =  0 --------(1)

6 x – 10 y – 40  =  0 --------(2)

From the above information, let us take the values of a1, a2, b1, b2, c1 and c2

a1  =  3, b1  =  -5, c1  =  -20

a1  =  6, b1  =  -10, c1  =  -40

a1/a =  3/6  =  1/2

b1/b =  -5/(-10)  =  1/2

c1/c2  =  (-20)/(-40)  =  1/2

here, a1/a2  =  b1/b2  =  c1/c2

Therefore the two given lines are coincident.

(iv) x – 3y – 7  =  0

3x – 3y – 15  =  0

Solution :

From the above information, let us take the values of a1, a2, b1, b2, c1 and c2

a1  =  1, b1  =  -3, c1  =  -7

a1  =  3, b1  =  -3, c1  =  -15

a₁/a ₂ = 1/3

b₁/b ₂ = -3/(-3)  =  1

c₁/c ₂ = (-7)/(-15) = 7/15

here, a1/a2  ≠  b1/ b2

So, the given two lines are intersecting.

x/(45  - 21)  =  y/(-21 + 15)  =  1/(-3+9)

x/24  =  y/(-6)  =  1/6

x/24  =  1/6           y/(-6)  =  1/6

x  =  24/6               y  =  -6/6

x  =  4                       y  =  -1

After having gone through the stuff and examples,  we hope that the students would have understood, types of solutions of pair of linear equations in two variables.

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