Solve the following equations by trial and error method:

5p + 2 = 17

Dear Student

LHS = 5p + 2
By substituting the value of p = 0
Then,
LHS = 5p + 2
= (5 × 0) + 2
=0 +2
=2
By comparing LHS and RHS
2≠17
LHS≠RHS
Hence, the value of p = 0 is not a solution to the given equation.

Let, p = 1 LHS = 5p + 2

= (5 × 1) + 2 =5 +2
=7

By comparing LHS and RHS
7≠17
LHS≠RHS
Hence, the value of p = 1 is not a solution to the given equation.

Let, p = 2
LHS = 5p + 2
= (5 × 2) + 2 = 10 + 2
= 12
By comparing LHS and RHS
12≠17
LHS≠RHS
Hence, the value of p = 2 is not a solution to the given equation.
Let, p = 3
LHS = 5p + 2
= (5 × 3) + 2 = 15 + 2
= 17
By comparing LHS and RHS 17 = 17
LHS = RHS

Thanks