Ex 3.5, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

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 x – 3y – 3 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 1, b1 = –3, c1 = –3 3x – 9y – 2 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 3, b2 = –9, c2 = –2 a1 = 1, b1 = –3, c1 = –3 & a2 = 3, b2 = –9, c2 = –2 𝒂𝟏/𝒂𝟐 𝑎1/𝑎2 = 1/3 𝒃𝟏/𝒃𝟐 𝑏1/𝑏2 = (−3)/(−9) 𝑏1/𝑏2 = 1/3 𝒄𝟏/𝒄𝟐 𝑐1/𝑐2 = (−3)/(−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 2x + y = 5 2x + 1y – 5 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 2, b1 = 1, c1 = –5 3x + 2y = 8 3x + 2y – 8 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 3, b2 = 2, c2 = –8 a1 = 2, b1 = 1, c1 = –5 & a2 = 3, b2 = 2, c2 = –8 𝒂𝟏/𝒂𝟐 𝑎1/𝑎2 = 2/3 𝒃𝟏/𝒃𝟐 𝑏1/𝑏2 = 1/2 𝒄𝟏/𝒄𝟐 𝑐1/𝑐2 = (−5)/(−8) 𝑐1/𝑐2 = 5/8 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 𝑥/(1×(−8) − 2 ×(−5) ) 𝑥/((−8) + 10 ) 𝑥/(2 ) = 𝑦/(3 ×(−5) − 2 × (−8) ) = 𝑦/(−15 + 16) = 𝑦/(1 ) = 1/(2 × 2 − 3 × 1 ) = 1/(4 − 3 ) = 1/1 Now, 𝒙/𝟐 = 𝟏/𝟏 x = 2 × 1 ∴ x = 2 𝒚/𝟏 = 𝟏/𝟏 y = 1 × 1 ∴ y = 1 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 3x – 5y = 20 3x – 5y – 20 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 3,b1 = –5,c1 = –20 6x – 10y = 40 6x – 10y – 40 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 6,b2 = –10,c2 = –40 a1 = 3, b1 = –5, c1 = –20 & a2 = 6, b2 = –10, c2 = –40 𝒂𝟏/𝒂𝟐 𝑎1/𝑎2 = 3/6 𝑎1/𝑎2 = 1/2 𝒃𝟏/𝒃𝟐 𝑏1/𝑏2 = (−5)/(−10) 𝑏1/𝑏2 = 1/2 𝒄𝟏/𝒄𝟐 𝑐1/𝑐2 = (−20)/(−40) 𝑐1/𝑐2 = 1/2 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 x – 3y – 7 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 1, b1 = –3, c1 = –7 3x – 3y – 15 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 3, b2 = –3, c2 = –15 a1 = 1, b1 = –3, c1 = –7 & a2 = 3, b2 = –3, c2 = –15 𝒂𝟏/𝒂𝟐 𝑎1/𝑎2 = 1/3 𝒃𝟏/𝒃𝟐 𝑏1/𝑏2 = (−3)/(−3) 𝑏1/𝑏2 = 1 𝒄𝟏/𝒄𝟐 𝑐1/𝑐2 = (−7)/(−15) 𝑐1/𝑐2 = 7/15 Since 𝒂𝟏/𝒂𝟐 ≠ 𝒃𝟏/𝒃𝟐 We have a unique solution Solving x – 3y – 7 = 0 …(1) 3x – 3y – 15 = 0 …(2) Using cross-multiplication 𝑥/(−3 ×(−15) − (−3) ×(−7) ) = 𝑦/(3 ×(−7) − 1 × (−15) ) = 1/(1 ×(−3) − 3 ×(−3) ) 𝑥/((45) − 21 ) = 𝑦/(−21 + 15 ) = 1/(−3 + 9 ) 𝑥/(24 ) = 𝑦/(−6 ) = 1/6 Now, 𝒙/𝟐𝟒 = 𝟏/𝟔 x = 24/6 ∴ x = 4 𝒚/(−𝟔) = 𝟏/𝟔 y = (−6)/6 ∴ y = – 1 Therefore, x = 4, y = –1 is the solution of our equation