1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method. (i)x – 3y – 3 = 0 ; 3x – 9y – 2 =0 (ii)2x + y = 5 ; 3x +2y =8 (iii)3x – 5y = 20 ; 6x – 10y =40 (iv)x – 3y – 7 = 0 ;3x – 3y – 15= 0

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

(i)x – 3y – 3 = 0  ; 3x – 9y – 2 =0

 (ii)2x + y = 5 ; 3x +2y =8

(iii)3x – 5y = 20 ; 6x – 10y =40

 (iv)x – 3y – 7 = 0 ;3x – 3y – 15= 0

Answer

(i)

x – 3 y  - 3 =  0

3x – 9y  - 2 =  0

Compare with

            we get 

a1 = 1 ,           b1 = -3  ,        and     c1 = -3 

a2  = 3 ,          b2  = -9           and     c2 = -2

Hence

So system has no solutions

(ii)

2x + y  = 5

3x + 2y  = 8

2x + y  - 5  = 0

3x + 2y  - 8 = 0

Compare with

            we get 

a1 = 2 ,           b1 = 1  ,         and     c1 = -5 

a2  = 3 ,          b2  = 2            and     c2 = -8

Hence

So system has unique solutions

Use the formula of cross multiplication

Plug the values in this formula we get

Solve them we get

Answer x = 2 and y = 1

(iii)

3x – 5y  = 20

6x – 10 y = 40

3x – 5y  -  20   = 0

6x – 10 y  - 40 = 0

Compare with

            we get 

a1 = 3 ,           b1 = -5  ,        and     c1 = -20 

a2  = 6 ,          b2  = -10         and     c2 = -40

Hence

So both lines are coincident and overlap with each other

So it will have infinity or many solutions

(iv)

X – 3y  - 7 = 0

3x – 3y  -15 = 0

Compare with

            we get 

a1 = 1 ,           b1 = -3  ,        and     c1 = -7 

a2  = 3 ,          b2  = -3           and     c2 = -15

Hence

So it will have unique solutions

Use the formula of cross multiplication

Plug the values in this formula we get

X = 24/6 = 4

Y = -6/6 = - 1

So our answer is x = 4 and y = - 1