Question -
Answer -
(i)
LHS = 5p + 2
By substituting thevalue of p = 0
Then,
LHS = 5p + 2
= (5 × 0) + 2
= 0 + 2
= 2
By comparing LHS andRHS
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 andRHS
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 andRHS
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 andRHS
17 = 17
LHS = RHS
Hence, the value of p= 3 is a solution to the given equation.
(ii)
LHS = 3m – 14
By substituting thevalue of m = 3
Then,
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing LHS andRHS
-5 ≠ 4
LHS ≠ RHS
Hence, the value of m= 3 is not a solution to the given equation.
Let, m = 4
LHS = 3m – 14
= (3 × 4) – 14
= 12 – 14
= – 2
By comparing LHS andRHS
-2 ≠ 4
LHS ≠ RHS
Hence, the value of m= 4 is not a solution to the given equation.
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing LHS andRHS
1 ≠ 4
LHS ≠ RHS
Hence, the value of m= 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing LHS andRHS
4 = 4
LHS = RHS
Hence, the value of m= 6 is a solution to the given equation.