10th cbse maths solution for exercise 3.5 part 1

This page 10th cbse maths solution for exercise 3.5 part 1is going to provide you solution for every problems that you find in the exercise no 3.5

10th CBSE maths solution for Exercise 3.5 part 1

(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 – 3 y – 3 = 0

     3 x – 9 y – 2 = 0

Solution:

From the above information let us take the values of a₁ , a₂, b₁, b₂, c₁ and c ₂

 a₁ = 1          b₁ = -3             c₁ = -3

 a₂ = 3         b ₂ = -9            c ₂ = -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)  2 x +y = 5

      3 x + 2 y = 8

Solution:

 2 x + y – 5 = 0

3 x + 2 y – 8 = 0

From the above information let us take the values of a₁ , a₂, b₁, b₂, c₁ and c ₂

 a₁ = 2         b₁ = 1            c₁ = -5

 a₂ = 3         b₂ = 2            c₂ = -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   10th CBSE maths solution for Exercise 3.5 part 1


(iii) 3 x – 5 y = 20

       6 x – 10 y = 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 c ₂

 a₁ = 3         b₁ = -5             c₁ = -20

 a₂ = 6         b₂ = -10            c₂ = -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 – 3 y – 7 = 0

     3 x – 3 y – 15 = 0

From the above information let us take the values of a₁ , a₂, b₁, b₂, c₁ and c ₂

 a₁ = 1         b₁ = -3              c₁= -7

 a ₂ = 3         b ₂ = -3            c ₂ = -15

a₁/a ₂ = 3/6 = 1/3

b₁/b ₂ = -3/(-3) = 1/1

c₁/c ₂ = (-7)/(-15) = 7/15

here, a₁/a ₂ ≠ b₁/b ₂

Therefore 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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