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
(i) x – 3y – 3 = 0
3x – 9y – 2 =0
a1/a2 = 1/3
b1/b2 = -3/-9 = 1/3 and
c1/c2 = -3/-2 = 3/2
a1/a2 = b1/b2 ≠ c1/c2
Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.
(ii) 2x + y = 5
3x +2y = 8
a1/a2 = 2/3
b1/b2 = 1/2 and
c1/c2 = -5/-8 = 5/8
a1/a2 ≠ b1/b2
Therefore, they will intersect each other at a unique point and thus, there will be 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 = 1, y/1 = 1
∴ x = 2, y = 1.
(iii) 3x – 5y = 20
6x – 10y = 40
a1/a2 = 3/6 = 1/2
b1/b2 = -5/-10 = 1/2 and
c1/c2 = -20/-40 = 1/2
a1/a2 = b1/b2 = c1/c2
Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.
(iv) x – 3y – 7 = 0
3x – 3y – 15= 0
a1/a2 = 1/3
b1/b2 = -3/-3 = 1 and
c1/c2 = -7/-15 = 7/15
a1/a2 ≠ b1/b2
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication,
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.