Question -
Answer -
(a)
LHS = n + 5
By substituting the value of n = 1
Then,
LHS = n + 5
= 1 + 5
= 6
By comparing LHS and RHS
6 тЙа 19
LHS тЙа RHS
Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.
(b)┬а
LHS = 7n + 5
By substituting the value of n = -2
Then,
LHS = 7n + 5
= (7 ├Ч (-2)) + 5
= тАУ 14 + 5
= тАУ 9
By comparing LHS and RHS
-9 тЙа 19
LHS тЙа RHS
Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.
(c)┬а
LHS = 7n + 5
By substituting the value of n = 2
Then,
LHS = 7n + 5
= (7 ├Ч (2)) + 5
= 14 + 5
= 19
By comparing LHS and RHS
19 = 19
LHS = RHS
Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.
(d)┬а
LHS = 4p тАУ 3
By substituting the value of p = 1
Then,
LHS = 4p тАУ 3
= (4 ├Ч 1) тАУ 3
= 4 тАУ 3
= 1
By comparing LHS and RHS
1 тЙа 13
LHS тЙа RHS
Hence, the value of p = 1 is not a solution to the given equation 4p тАУ 3 = 13.
(e)┬а
LHS = 4p тАУ 3
By substituting the value of p = тАУ 4
Then,
LHS = 4p тАУ 3
= (4 ├Ч (-4)) тАУ 3
= -16 тАУ 3
= -19
By comparing LHS and RHS
-19 тЙа 13
LHS тЙа RHS
Hence, the value of p = -4 is not a solution to the given equation 4p тАУ 3 = 13.
(f)┬а
LHS = 4p тАУ 3
By substituting the value of p = 0
Then,
LHS = 4p тАУ 3
= (4 ├Ч 0) тАУ 3
= 0 тАУ 3
= -3
By comparing LHS and RHS
тАУ 3 тЙа 13
LHS тЙа RHS
Hence, the value of p = 0 is not a solution to the given equation 4p тАУ 3 = 13.