Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

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  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise

Transcript

Ex 3.2, 3 On comparing the ratios ๐‘Ž1/๐‘Ž2 , ๐‘1/๐‘2 & ๐‘1/๐‘2 , find out whether the following pair of linear equations are consistent, or inconsistent. 3x+ 2y= 5 ; 2xโ€“ 3y= 7 3x + 2y โ€“ 5 = 0 2x โ€“ 3y โ€“ 7 = 0 3x + 2y โ€“ 5 = 0 Comparing with a1x + b1y + c1 = 0 โˆด a1 = 3 , b1 = 2 , c1 = โ€“5 2x โ€“ 3y โ€“ 7 = 0 Comparing with a2x + b2y + c2 = 0 โˆด a2 = 2 , b2 = โ€“3 , c2 = โ€“7 a1 = 3 , b1 = 2 , c1 = โ€“5 & a2 = 2 , b2 = โ€“3 , c2 = โ€“7 Since ๐‘Ž1/๐‘Ž2 โ‰  ๐‘1/๐‘2 We have a unique solution Therefore, our system is consistent. ๐’‚๐Ÿ/๐’‚๐Ÿ ๐’ƒ๐Ÿ/๐’ƒ๐Ÿ ๐’„๐Ÿ/๐’„๐Ÿ ๐‘Ž1/๐‘Ž2 = 3/2 ๐‘1/๐‘2 = 2/(โˆ’3) ๐‘1/๐‘2 = -2/3 ๐‘1/๐‘2 = (โˆ’5)/(โˆ’7) ๐‘1/๐‘2 = 5/7 Ex 3.2, 3 On comparing the ratios ๐‘Ž1/๐‘Ž2 , ๐‘1/๐‘2 & ๐‘1/๐‘2 , find out whether the following pair of linear equations are consistent, or inconsistent. (ii) 2x -3y = 8 ; 4x - 6y = 9 2x โ€“ 3y โˆ’ 8 = 0 4x โ€“ 6y โ€“ 9 = 0 2x โ€“ 3y โˆ’ 8 = 0 Comparing with a1x + b1y + c1 = 0 โˆด a1 = 2 , b1 = โˆ’ 3 , c1 = โ€“ 8 4x โ€“ 6y โ€“ 9 = 0 Comparing with a2x + b2y + c2 = 0 โˆด a2 = 4 , b2 = โ€“ 6 , c2 = โ€“ 9 โˆด a1 = 2, b1 = โˆ’ 3 , c1 = โˆ’ 8 a2 = 4, b2 = โˆ’6, c2 = โˆ’ 9 Since ๐‘Ž1/๐‘Ž2 = ๐‘1/๐‘2 โ‰  ๐‘1/๐‘2 We have no solution. Therefore, our system is inconsistent. ๐‘Ž1/๐‘Ž2 = 2/4 ๐‘Ž1/๐‘Ž2 = 1/2 ๐‘1/๐‘2 = (โˆ’3)/(โˆ’6) ๐‘1/๐‘2 = 1/2 ๐‘1/๐‘2 = (โˆ’8)/(โˆ’9) ๐‘1/๐‘2 = 8/9 ๐‘1/๐‘2 = (โˆ’8)/(โˆ’9) ๐‘1/๐‘2 = 8/9 Ex 3.2, 3 On comparing the ratios ๐‘Ž1/๐‘Ž2 , ๐‘1/๐‘2 & ๐‘1/๐‘2 , find out whether the following pair of linear equations are consistent, or inconsistent. (iii) 3/2x + 5/3y = 7 ; 9x - 10y = 14 3/2 ๐‘ฅ + 5/3 ๐‘ฆ โ€“ 7 =0 9x โ€“ 10y โ€“ 14 = 0 ๐Ÿ‘/๐Ÿ ๐’™ + ๐Ÿ“/๐Ÿ‘ ๐’š โ€“ ๐Ÿ• =๐ŸŽ Comparing with a1x + b1y + c1 = 0 โˆด a1 = 3/2 , b1 = 5/3 , c1 = โ€“ 7 9x โ€“ 10y โ€“ 14 = 0 Comparing with a2x + b2y + c2 = 0 โˆด a2 = 9 , b2 = โ€“ 10, c2 = โ€“ 14 ๐‘Ž1/๐‘Ž2 = (3/2)/9 ๐‘Ž1/๐‘Ž2 = 3/(2 ร— 9) ๐‘Ž1/๐‘Ž2 = 1/(2 ๐‘Žร—3) ๐‘Ž1/๐‘Ž2 = 1/6 ๐‘1/๐‘2 = (5/3)/(โˆ’10) ๐‘1/๐‘2 = 5/(โˆ’3 ร—10) ๐‘1/๐‘2 = 1/(โˆ’3 ร— 2) ๐‘1/๐‘2 = (โˆ’1)/6 ๐‘1/๐‘2 = (โˆ’7)/(โˆ’14) ๐‘1/๐‘2 = 1/2 Since ๐‘Ž1/๐‘Ž2 โ‰  ๐‘1/๐‘2 We have a unique solution Therefore, our system is consistent. Ex 3.2,3 On comparing the ratios ๐‘Ž1/๐‘Ž2 , ๐‘1/๐‘2 & ๐‘1/๐‘2 , find out whether the following pair of linear equations are consistent, or inconsistent. (iv) 5x โ€“ 3y = 11 , -10x + 6y = -22 5x โ€“ 3y โ€“ 11 = 0 โˆ’10x + 6y + 22 = 0 5x โ€“ 3y โ€“ 11 = 0 Comparing with a1x + b1y + c1 = 0 โˆด a1 = 5 , b1 = โˆ’3 , c1 = โ€“ 11 โˆ’10x + 6y + 22 = 0 Comparing with a2x + b2y + c2 = 0 โˆด a2 = โˆ’ 10 , b2 = 6, c2 = 22 โˆด a1 = 5 , b1 = โˆ’3 , c1 = โ€“ 11 & a2 = โˆ’ 10 , b2 = 6, c2 = 22 Since ๐‘Ž1/๐‘Ž2 = ๐‘1/๐‘2 = ๐‘1/๐‘2 We have infinitely many solutions Therefore, our system is consistent. ๐’‚๐Ÿ/๐’‚๐Ÿ ๐‘Ž1/๐‘Ž2 = 5/(โˆ’10) ๐‘Ž1/๐‘Ž2 = โˆ’1/2 ๐‘1/๐‘2 = (โˆ’3)/6 ๐‘1/๐‘2 = โˆ’1/2 ๐‘1/๐‘2 = (โˆ’11)/22 ๐‘1/๐‘2 = โˆ’ 1/2 โˆด a1 = 5 , b1 = โˆ’3 , c1 = โ€“ 11 & a2 = โˆ’ 10 , b2 = 6, c2 = 22 Since ๐‘Ž1/๐‘Ž2 = ๐‘1/๐‘2 = ๐‘1/๐‘2 We have infinitely many solutions Therefore, our system is consistent. Ex 3.2, 3 On comparing the ratios ๐‘Ž1/๐‘Ž2 , ๐‘1/๐‘2 & ๐‘1/๐‘2 , find out whether the following pair of linear equations are consistent, or inconsistent. (v) 4/3 x + 2y = 8 ; 2x + 3y = 12 4/3 x + 2y โ€“ 8 = 0 2x + 3y โ€“ 12 = 0 ๐Ÿ’/๐Ÿ‘ x + 2y โ€“ 8 = 0 Comparing with a1x + b1y + c1 = 0 โˆด a1 = 4/3 , b1 = 2 , c1 = โ€“ 8 2x + 3y โ€“ 12 = 0 Comparing with a2x + b2y + c2 = 0 โˆด a2 = 2, b2 = 3 c2 = โˆ’12 ๐‘Ž1/๐‘Ž2 = (4/3)/2 ๐‘Ž1/๐‘Ž2 = 4/(3 ร— 2) ๐‘Ž1/๐‘Ž2 = 2/(3 ร— 1) ๐‘Ž1/๐‘Ž2 = 2/3 ๐‘1/๐‘2 = 2/3 ๐‘1/๐‘2 = (โˆ’8)/(โˆ’12) ๐‘1/๐‘2 = 2/3 Since ๐‘Ž1/๐‘Ž2 = ๐‘1/๐‘2 = ๐‘1/๐‘2 We have infinite solutions. Therefore, our system is consistent.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.