Question 2 - Solving difficult equations - Chapter 2 Class 8 Linear Equations in One Variable
Last updated at April 16, 2024 by Teachoo
Solving difficult equations
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Solving difficult equations
Last updated at April 16, 2024 by Teachoo
Question 2 A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?Let the other number be x Given that, A positive number is 5 times that number ∴ Positive number = 5x Now, 21 added to both numbers, ∴ New other number = x + 21 New positive number = 5x + 21 Given that, One of the new number becomes twice of the other new number 5x + 21 = 2 (x + 21) 5x + 21 = 2x + 42 5x − 2x + 21 = 42 3x + 21 = 42 3x = 42 − 21 3x = 21 x = 21/3 x = 7 ∴ Another number is x = 7 Positive number is 5x = 5 × 7 = 35 Rough Here, First new number = x + 21 Second number = 5x + 21 So, second number is greater Second number will be twice of first ∴ It will be 5x + 21 = 2 (x + 21)