Question

3. Solve the following equations by trial and error method:
(1) 5p + 2 = 17
(i) 3m - 14 = 4​

Answer 1

Answer:

1.5p + 2 = 17

Put p = 1 in LHS

5 * 1 + 2 = 5 + 2 = 7 ≠ RHS

Put p = 2 in LHS

5 * 2 + 2 = 10 + 2 = 12 ≠ RHS

Put p = 3 in LHS

5 * 3 + 2 = 15 + 2 = 17 = RHS

Hence, p = 3 is the solution of the given method.

(ii) 3m – 14 = 4

Put m = 4 in LHS

3 * 4 – 14 = 12 – 14 = -2 ≠ RHS

Put m = 5 in LHS

3 * 5 – 14 = 15 – 14 = 1 ≠ RHS

Put m = 6 in LHS

3 * 6 – 14 = 18 – 14 = 4 = RHS

Hence, m = 6 is the solution of the given method.

Hope it helps u.. ☺️☺️✌️✌️

Answer 2

Answer:

$\bf(i)5p + 2 = 17 \\ \bf5p = 17 - 2 \\ \ \bf p = \frac{15}{5} \\ \bf \: p = 3 \\ \\ \\ \bf (iii)3m - 14 = 4 \\ \bf3m = 4 + 14 \\ \bf \: m = \frac{18}{3} \\ \bf \: m = 6$

hope it helps you dear....