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
a₁ x + b₁ y + c₁ = 0
a₂ x + b₂ y + c₂ = 0
a) a₁/a₂ ≠ b₁/b₂ (Intersecting Lines)
b) a₁/a₂ = b₁/b₂ = c₁/c₂ (Coincident Lines)
c) a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (Parallel Lines)
(i) x - 3y - 3 = 0; 3x - 9 y - 2 = 0
a₁/a₂= 1/3
b₁/b₂= -3/-9 = 1/3
c₁/c₂= -3/-2 = 3/2
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
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
a₁/a₂= 2/3
b₁/b₂= 1/2
c₁/c₂= -5/-8 = 5/8
a₁/a₂ ≠ b₁/b₂
Therefore, A unique solution for these equations.
By cross-multiplication method,
[x/(b₁c₂- b₂c₁) = y/(c₁a₂- c₂a₁) = 1/(a₁b₂ - a₂b₁)]
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
a₁/a₂= 3/6 = 1/2
b₁/b₂= 5/10 = 1/2
c₁/c₂= -20/-40 = 1/2
a₁/a₂ = b₁/b₂ = c₁/c₂
Therefore, Infinite solutions possible for these equations.
(iv) x - 3y - 7 = 0; 3x - 3y - 15 = 0
a₁/a₂ = 1/3
b₁/b₂ = -3/-3 = 1
c₁/c₂ = -7/-15 = 7/15
a₁/a₂ ≠ b₁/b₂
Therefore, A unique solution for these equations.
By cross-multiplication method,
[x/(b₁c₂- b₂c₁) = y/(c₁a₂- c₂a₁)-= 1/(a₁b₂ - a₂b₁)]
[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
☛ Check: NCERT Solutions for Class 10 Maths Chapter 3
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
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
☛ Related Questions:
- (i) For which values of a and b will the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7 (a - b) x + (a + b) y = 3a + b - 2 (ii) For which value of k will the following pair of linear equations have no solution? 3x + y = 1 (2k -1) x + (k -1) y = 2k +1
- Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5 y = 9 3x + 2 y = 4.
- Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
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