Ex 3.5, 1 - Class 10

Transcript

 Ex 3.5 ,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 x – 3y – 3 = 0 3x – 9y – 2 = 0 x – 3y – 3 = 0 3x – 9y – 2 = 0 a1 = 1 , b1 = –3 , c1 = –3 & a2 = 3 , b2 = –9 , c2 = –2 Since 𝑎1/𝑎2 = 𝑏1/𝑏2 ≠ 𝑐1/𝑐2 We have no solution.  Ex 3.5 ,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 (ii) 2x + y = 5 3x + 2y = 8 2x + y = 5 3x + 2y = 8 a1 = 2 , b1 = 1 , c1 = –5 & a2 = 3 , b2 = –9 , c2 = –2 Since 𝑎1/𝑎2 ≠ 𝑏1/𝑏2 We have a unique solution Solving 2x + y = 5 …(1) 3x + 2y = 8 …(2) For cross-multiplication 2x + y – 5 = 0 3x + 2y – 8 = 0 Now, 𝑥/2 = 1/1 x = 2 × 1 ∴ x = 2 Therefore, x = 2 , y = 1 is the solution of our equation  Ex 3.5 ,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 (iii) 3x – 5y = 20 6x – 10y = 40 3x – 5y = 20 6x – 10y = 40 a1 = 3 , b1 = –5 , c1 = –20 & a2 = 6 , b2 = –10 , c2 = –40 Since 𝑎1/𝑎2 = 𝑏1/𝑏2 = 𝑐1/𝑐2 We have infinitely many solutions Ex 3.5 ,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 (iv) x – 3y – 7 = 0 3x – 3y – 15 = 0 x – 3y – 7 = 0 3x – 3y – 15 = 0 a1 = 1 , b1 = –3 , c1 = –7 & a2 = 3 , b2 = –3 , c2 = –15 Since 𝑎1/𝑎2 ≠ 𝑏1/𝑏2 We have a unique solution Solving x – 3y – 7 = 0 …(1) 3x – 3y – 15 = 0 …(2) Using cross-multiplication Now, 𝑥/24 = 1/6 x = 24/6 ∴ x = 4 Therefore, x = 4 , y = –1 is the solution of our equation