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 ≠ 19
LHS ≠ RHS
Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.
(b)
LHS = 7n + 5
By substituting the value of n = -2
Then,
LHS = 7n + 5
= (7 × (-2)) + 5
= – 14 + 5
= – 9
By comparing LHS and RHS
-9 ≠ 19
LHS ≠ RHS
Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.
(c)
LHS = 7n + 5
By substituting the value of n = 2
Then,
LHS = 7n + 5
= (7 × (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)
LHS = 4p – 3
By substituting the value of p = 1
Then,
LHS = 4p – 3
= (4 × 1) – 3
= 4 – 3
= 1
By comparing LHS and RHS
1 ≠ 13
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.
(e)
LHS = 4p – 3
By substituting the value of p = – 4
Then,
LHS = 4p – 3
= (4 × (-4)) – 3
= -16 – 3
= -19
By comparing LHS and RHS
-19 ≠ 13
LHS ≠ RHS
Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.
(f)
LHS = 4p – 3
By substituting the value of p = 0
Then,
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= -3
By comparing LHS and RHS
– 3 ≠ 13
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.