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Ex 4.2, 3 Check which of the following are solutions of the equation π‘₯βˆ’2𝑦=4 and which are not: (iii) (4, 0) Given equation π‘₯βˆ’2𝑦=4 Putting x = 4 and y = 0 in LHS π’™βˆ’πŸπ’š =4βˆ’2(0) =πŸ’ Since, LHS = RHS Hence, (4, 0) is a solution

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.