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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.