Question 2 exercise 4.1

(e) 4p – 3 = 13; (p = -A)
Put p = -4 in LHS
4 × (-4) – 3 = -16 – 3 = -19 ≠ 13 (RHS)
Since LHS ≠ RHS
Thus p = -4 is not the solution of the given equation.

(f) 4p – 3 = 13; (p = 0)
Put p = 0 in LHS
4 × (0) – 3 = 0 – 3 = -3 ≠ 13 (RHS)
Since LHS ≠ RHS
Thus p – 0 is not the solution of the given equation.

Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19; (n = 1)
(b) 7n + 5 = 19; in – -2)
(c) 7n + 5 = 19; (n = 2)
(d) 4p – 3 = 13; (p = 1)
(e) 4p – 3 = 13; (p = -4)
(f) 4p-3 = 13; (p = 0)
Solution:
(a) n + 5 = 19 (n = 1)
Put n = 1 in LHS
1 + 5 = 6 ≠ 19 (RHS)
Since LHS ≠ RHS
Thus n = 1 is not the solution of the given equation.

(b) 7n + 5 = 19; (n = -2)
Put n = – 2 in LHS
7 × (-2) + 5 = -14 + 5 = -9 ≠ 19 (RHS)
Since LHS ≠ RHS
Thus, n = -2 is not the solution of the given equation.

(c) 7n+ 5 = 19; (n = 2)
Put n = 2 in LHS
7 × 2 + 5 = 14 + 5 = 19 = 19 (RHS)
Since LHS = RHS
Thus, n – 2 is the solution of the given equation.

(d) 4p – 3 = 13; (p = 1)
Put p = 1 in LHS
4 × 1 – 3 = 4 – 3 = 1 ≠ 13 (RHS)
Since LHS ≠ RHS
Thus, p = 1 is not the solution of the given equation.