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.

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/b₂  =  -3/(-9)  =  1/3

c₁/c₂  =  -3/(-2)  =  3/2

Here a₁/a₂  =  b₁/b₂  ≠  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/a₂  =  2/3

b₁/b₂  =  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/a₂  =  3/6  =  1/2

b₁/b₂  =  -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  ≠  b_1/ b₂

So, the given two lines are intersecting.

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

x/24  =  y/(-6)  =  1/6

Example 2 :

The pair of linear equations 2𝑥 = 5y + 6 and 15y = 6𝑥 - 8 represents two lines which are

(a)  Intersecting    (b) Parallel   (c) coincident

(d) either Intersecting or Parallel

Solution :

2𝑥 = 5y + 6 and 15y = 6𝑥 - 8

By comparing slope and y-intercept of these two lines, we can easily decide what type of lines these are.

Since both lines are having same slope and different y-intercepts, the lines must be parallel and they will not intersect each other. Then they will not have solution. So, option b parallel is correct.

Example 3 :

If the pair of linear equations 𝑥 - y = 1, 𝑥 + ky = 5 has a unique solution 𝑥 = 2, y = 1 then the value of k is

(a) -2    (b) -3    (c) 3     (d) 4

Solution :

𝑥 - y = 1 ------(1)

𝑥 + ky = 5 ------(2)

Here x = 2 and y = 1 is solution, by applying these values in (2), we get

2 + k(1) = 5

2 + k = 5

k = 5 - 2

k = 3

So, option c is correct.

Example 4 :

The pair of linear equations 3𝑥 + 5y = 3 and 6𝑥 + ky = 8 do not have a solution if k

(a) = 5   (b) = 10   (c) ≠10    (d) ≠ 5

Solution :

3𝑥 + 5y = 3 and 6𝑥 + ky = 8

Given that the lines do not have solution, then they must be parallel lines.

For parallel lines, slope will be equal and y-intercept will not be equal.

Equating the slopes, 

-3/5 = -6/k

k = 6(5/3)

k = 2(5)

k = 10

So, option b is correct.

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