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.

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

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

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