Check sibling questions


Transcript

Ex 4.1, 2 Check whether the value given in the brackets is a solution to the given equation or not: (a) n + 5 = 19 (n = 1) n + 5 = 19 Putting n = 1 1 + 5 = 19 6 = 19 It is not true. Hence, n = 1 is not a solution. Ex 4.1, 2 Check whether the value given in the brackets is a solution to the given equation or not: (b) 7n + 5 = 19 (n = – 2) 7n + 5 = 19 Putting n = −2 7 (−2) + 5 = 19 −14 + 5 = 19 −(14 − 5) = 19 −9 = 19 It is not true. Hence, n = −2 is not a solution. Ex 4.1, 2 Check whether the value given in the brackets is a solution to the given equation or not: (c) 7n + 5 = 19 (n = 2) 7n + 5 = 19 Putting n = 2 7 (2) + 5 = 19 14 + 5 = 19 19 = 19 It is true. Hence, n = 2 is a solution. Ex 4.1, 2 Check whether the value given in the brackets is a solution to the given equation or not: (d) 4p – 3 = 13 (p = 1) 4p − 3 = 13 Putting p = 1 4(1) + 3 = 13 4 − 3 = 13 1 = 13 It is not true. Hence, p = 1 is not a solution. Ex 4.1, 2 Check whether the value given in the brackets is a solution to the given equation or not: (e) 4p – 3 = 13 (p = – 4) 4p − 3 = 13 Putting p = −4 4(−4) − 3 = 13 −16 − 3 = 13 −(16 + 3) = 13 −19 = 13 It is not true. Hence, p = −4 is not a solution. Ex 4.1, 2 Check whether the value given in the brackets is a solution to the given equation or not: (f) 4p – 3 = 13 (p = 0) 4p − 3 = 13 Putting p = 0 4(0) − 3 = 13 0 − 3 = 13 −3 = 13 It is not true. Hence, p = 0 is not a solution.

  1. Chapter 4 Class 7 Simple Equations
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo