Question -
Answer -
(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8
(iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0
Solutions:
(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0
a1/a2=1/3 , b1/b2= -3/-9 =1/3, c1/c2=-3/-2 = 3/2
(a1/a2) = (b1/b2) ≠ (c1/c2)
Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.
(ii) Given, 2x + y = 5 and 3x +2y = 8
a1/a2 = 2/3 , b1/b2 = 1/2 , c1/c2 = -5/-8
(a1/a2) ≠ (b1/b2)
Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:
x/(b1c2-c1b2) = y/(c1a2 – c2a=) = 1/(a1b2-a2b1)
x/(-8-(-10)) = y/(15+16) = 1/(4-3)
x/2 = y/1 = 1
∴ x = 2 and y =1
(iii) Given, 3x – 5y = 20 and 6x – 10y = 40
(a1/a2) = 3/6 = 1/2
(b1/b2) = -5/-10 = 1/2
(c1/c2) = 20/40 = 1/2
a1/a2 = b1/b2 = c1/c2
Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.
(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0
(a1/a2) = 1/3
(b1/b2) = -3/-3 = 1
(c1/c2) = -7/-15
a1/a2 ≠ b1/b2
Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.
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.