Solve the following equations by trial and error method: 5p+2 =17

5p+2 =17
Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13
∵ -13 ≠ 17 Therefore, p = -3 is not the solution.

Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8
∵ -8 ≠ 17 Therefore, p = -2 is not the solution.

Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3
∵ -3 ≠ 17 Therefore, p = -1 is not the solution.

Putting p = 0 in L.H.S. 5(0)+2 = 0+2 = 2
∵ 2 ≠ 17 Therefore, p = 0 is not the solution.

Putting p = 1 in L.H.S. 5(1)+2 = 5+2 = 7
∵ 7 ≠ 17 Therefore, p = 1 is not the solution.

Putting p = 2 in L.H.S. 5(2)+2 = 10+2 = 12
∵ 12 ≠ 17 Therefore, p = 2 is not the solution.

Putting p = 3 in L.H.S. 5(3)+2 = 15+2 = 17
∵ 17 ≠ 17 Therefore, p = 3 is the solution.

Class 7 Maths Chapter 4 Exercise 4.1

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/