Q.2 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)

(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)

(a) n + 5 = 19 (n = 1)
Put n = 1 in LHS

1 + 5 = 6

RHS =19

Since LHS ? RHS

Thus n = 1 is not the solution of the given equation. "n+5=19"

(b) 7n + 5 = 19 (n = -2)
Put n = -2 in LHS

-2×7 + 5 = -9

RHS =19

Since LHS \(\neq \)RHS

Thus n = -2 is not the solution of the given equation. "7n+5=19"

(c) 7n + 5 = 19 (n = 2)
Put n = 2 in LHS

7×2 + 5 = 19

RHS =19

Since LHS = RHS

Thus n = 2 is the solution of the given equation. "7n+5=19"

(d) 4p-3 = 13 (p= 1)
Put p = 1 in LHS

4×1-3 = 1

RHS =13

Since LHS \(\neq \)RHS

Thus p = 1 is not the solution of the given equation. "4p-3=13"

(e) 4p - 3= 13 (p = -4)
Put p = -4 in LHS

4×(-4) - 3 = -19

RHS =13

Since LHS \(\neq \)RHS

Thus p = -4 is not the solution of the given equation. "4p-3=13"

(f) 4p-3=13 (p= 0)
Put p = 0 in LHS

0×4-3 = -3

RHS =13

Since LHS \(\neq \) RHS

Thus p = 0 is not the solution of the given equation. "4p-3=13"