# 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 the 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

**Solution:**

For any pair of linear equation

a_{1} x + b_{1} y + c_{1} = 0

a_{2} x + b_{2} y + c_{2} = 0

a) a_{1}/a_{2} ≠ b_{1}/b_{2} (Intersecting Lines)

b) a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2} (Coincident Lines)

c) a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2} (Parallel Lines)

(i) x - 3y - 3 = 0; 3x - 9 y - 2 = 0

a_{1}/a_{2}= 1/3

b_{1}/b_{2}= -3/-9 = 1/3

c_{1}/c_{2}= -3/-2 = 3/2

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Therefore, In the given problem the given sets of lines are parallel and no solution for these equations.

(ii) 2x + y = 5; 3x + 2y = 8

2x + y - 5 = 0

3x + 2 y - 8 = 0

a_{1}/a_{2}= 2/3

b_{1}/b_{2}= 1/2

c_{1}/c_{2}= -5/-8 = 5/8

a_{1}/a_{2} ≠ b_{1}/b_{2}

Therefore, A unique solution for these equations.

By cross-multiplication method,

[x/(b_{1}c_{2}- b_{2}c_{1}) = y/(c_{1}a_{2}- c_{2}a_{1}) = 1/(a_{1}b_{2} - a_{2}b_{1})]

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

x/2 = y/1 = 1

∴ x = 2 and y = 1

(iii) 3x - 5 y = 20; 6x - 10 y = 40

3x - 5y - 20 = 0

6x -10 y - 40 = 0

a_{1}/a_{2}= 3/6 = 1/2

b_{1}/b_{2}= 5/10 = 1/2

c_{1}/c_{2}= -20/-40 = 1/2

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Therefore, Infinite solutions possible for these equations.

(iv) x - 3y - 7 = 0; 3x - 3y - 15 = 0

a_{1}/a_{2 }= 1/3

b_{1}/b_{2 }= -3/-3 = 1

c_{1}/c_{2 }= -7/-15 = 7/15

a_{1}/a_{2} ≠ b_{1}/b_{2}

Therefore, A unique solution for these equations.

By cross-multiplication method,

[x/(b_{1}c_{2}- b_{2}c_{1}) = y/(c_{1}a_{2}- c_{2}a_{1})-= 1/(a_{1}b_{2} - a_{2}b_{1})]

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

x/24 = y/-6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

x = 4 and y = - 1

∴ x = 4, y = -1

**Video Solution:**

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

### NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.5 Question 1:

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

The pair of equations that have unique solutions are ( 2x + y = 5; 3x + 2y = 8 ) and (x - 3y - 7 = 0; 3x - 3y - 15 = 0) no solutions are x - 3y - 3 = 0; 3x - 9 y - 2 = 0 infinitely many solutions are 3x - 5 y = 20; 6x - 10 y = 40