Example 9 - Let us consider Example 2 in Section 3.3 - Examples

Example 9 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Example 9 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3 Example 9 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4

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

Transcript

Example 9 Let us consider Example 2 in Section 3.3, i.e., the cost of 2 pencils and 3 erasers is Rs 9 and the cost of 4 pencils and 6 erasers is Rs 18. Find the cost of each pencil and each eraser. We formed the equations in Example 2 Our equations are 2x + 3y = 9 ...(1) 4x + 6y = 18 …(2) From (1) 2x + 3y = 9 2x = 9 – 3y x = ((𝟗 − 𝟑𝒚)/𝟐) Putting value of x in (2) 4x + 6y = 18 4 ((9 − 3𝑦)/2)+6𝑦=18 2 (9 – 3y) + 6y = 18 18 – 6y + 6y = 18 18 = 18 It is true for all values of x & y. Since the given equations have infinitely many solutions, That is why we cannot obtain a unique solution Reason :- The 2 equations given in question are 2x + 3y = 9 …(1) 4x + 6y = 18 …(2) From (2), 4x + 6y = 18 Diving both sides by 2 4𝑥/2 + 6𝑦/2 = 18/2 2x + 3y = 9 Which is equation same as equation (1) Hence both equation actually same So there can be infinite values of x and y For Example If y = 1, 2x + 3(1) = 9 2x + 3 = 9 2x = 9 – 3 2x = 6 x = 6/2 x = 3 If y = 0, 2x + 3(0) =9 2x + 0 = 9 2x = 9 x = 9/2 And so on

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.