\\\\(x\\\\)\\\\(0\\\\) \\\\( 1\\\\)\\\\( 2\\\\) \\\\(y\\\\)\\\\(3\\\\)\\\\( 8\\\\) \\\\( 13\\\\) Taxi fare for first kilometer \\\\(=\\\\) Rs. \\\\(8\\\\)Taxi fare for subsequent distance \\\\(=\\\\) Rs. \\\\(5\\\\)Total distance covered \\\\(= x\\\\)Total fare \\\\(= y\\\\)Since the fare for first kilometer \\\\(=\\\\) Rs.\\\\( 8\\\\)According to problem,Fare for \\\\((x – 1)\\\\) kilometer \\\\(=\\\\) \\\\(5(x-1)\\\\)So, the total fare \\\\(y = 5(x-1) + 8\\\\)\\\\(\\Rightarrow y = 5(x-1) + 8\\\\)\\\\(\\Rightarrow y = 5x – 5 + 8\\\\)\\\\(\\Rightarrow y = 5x + 3\\\\)\\\\(Hence\\\\), \\\\(y = 5x + 3\\\\) \\\\(\ ext{is the required linear equation.}\\\\)Now the equation is\\\\(y = 5x + 3\\\\) ...(1)Now, putting the value \\\\(x = 0\\\\) in (1)\\\\(y = 5 \ imes 0 + 3\\\\)\\\\(y = 0 + 3 = 3 \\\\) So the solution is \\\\((0, 3)\\\\)Putting the value \\\\(x = 1\\\\) in (1)\\\\(y = 5 \ imes 1 + 3\\\\)\\\\(y = 5 + 3 = 8\\\\). So the solution is \\\\((1, 8)\\\\)Putting the value \\\\(x = 2\\\\) in (1)\\\\(y = 5 \ imes 2 + 3\\\\)\\\\(y = 10 + 3 = 13\\\\). So the solution is \\\\((2, 13)\\\\)