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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 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

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 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

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 60787 Points
2 years ago
(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.

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