# Q.3 Solve the following equations by trial and error method: (i) 5p + 2 = 17 (ii) 3m – 14 = 4

(i) 5p + 2 = 17
For p = 1, LHS
= 5 × 1 + 2 = 5 + 2 = 7 $$\neq$$RHS=17
For p = 2, LHS = 5 × 2 + 2
= 10 + 2 = 12 $$\neq$$ RHS=17
For p = 3, LHS = 5 × 3 + 2
= 15 + 2 = 17 = RHS
Since the given equation is satisfied for p = 3 Thus, p = 3 is the required solution.
(ii) 3m – 14 = 4
For m = 1, LHS = 3 × 1 – 14
= 3 – 14 = -11 $$\neq$$ RHS = 4
For m = 2, LHS = 3 × 2 – 14 = 6 – 14 = -8 $$\neq$$ RHS= 4
For m = 3, LHS = 3 × 3 – 14 = 9 – 14 = -5 $$\neq$$ RHS =4
Form m = 4, LHS = 3 × 4 – 14 = 12 – 14 = -2 $$\neq$$ RHS= 4
For m = 5, LHS = 3 × 5 – 14 = 15 – 14 = -1 $$\neq$$ RHS =4
For m = 6, LHS = 3 × 6 – 14 = 18 – 14 = 4 =RHS
Since, the given equation is satisfied for m = 6. Thus, m = 6 is the required solution.