Question -
Answer -
(i)
LHS = 5p + 2
By substituting thevalue of p = 0
Then,
LHS = 5p + 2
= (5 ├Ч 0) + 2
= 0 + 2
= 2
By comparing LHS andRHS
2 тЙа 17
LHS тЙа RHS
Hence, the value of p= 0 is not a solution to the given equation.
Let, p = 1
LHS = 5p + 2
= (5 ├Ч 1) + 2
= 5 + 2
= 7
By comparing LHS andRHS
7 тЙа 17
LHS тЙа RHS
Hence, the value of p= 1 is not a solution to the given equation.
Let, p = 2
LHS = 5p + 2
= (5 ├Ч 2) + 2
= 10 + 2
= 12
By comparing LHS andRHS
12 тЙа 17
LHS тЙа RHS
Hence, the value of p= 2 is not a solution to the given equation.
Let, p = 3
LHS = 5p + 2
= (5 ├Ч 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 тАУ 14
By substituting thevalue of m = 3
Then,
LHS = 3m тАУ 14
= (3 ├Ч 3) тАУ 14
= 9 тАУ 14
= тАУ 5
By comparing LHS andRHS
-5 тЙа 4
LHS тЙа RHS
Hence, the value of m= 3 is not a solution to the given equation.
Let, m = 4
LHS = 3m тАУ 14
= (3 ├Ч 4) тАУ 14
= 12 тАУ 14
= тАУ 2
By comparing LHS andRHS
-2 тЙа 4
LHS тЙа RHS
Hence, the value of m= 4 is not a solution to the given equation.
Let, m = 5
LHS = 3m тАУ 14
= (3 ├Ч 5) тАУ 14
= 15 тАУ 14
= 1
By comparing LHS andRHS
1 тЙа 4
LHS тЙа RHS
Hence, the value of m= 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m тАУ 14
= (3 ├Ч 6) тАУ 14
= 18 тАУ 14
= 4
By comparing LHS andRHS
4 = 4
LHS = RHS
Hence, the value of m= 6 is a solution to the given equation.