Page 3 - Lesson Note- Equation to Statement and Statement to Equation Ch-4-converted

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