Question

Solve the following equations by trial and error method:

(i) 5p + 2 = 17 (ii) 3m − 14 = 4

Answer 1

(i) 5p + 2 = 17

Putting p = 1 in L.H.S.,

(5 × 1) + 2 = 7 ≠ R.H.S.

Putting p = 2 in L.H.S.,

(5 × 2) + 2 = 10 + 2 = 12 ≠ R.H.S.

Putting p = 3 in L.H.S.,

(5 × 3) + 2 = 17 = R.H.S.

Hence, p = 3 is a solution of the given equation.

(ii) 3m − 14 = 4

Putting m = 4,

(3 × 4) − 14 = −2 ≠ R.H.S.

Putting m = 5,

(3 × 5) − 14 = 1 ≠ R.H.S.

Putting m = 6,

(3 × 6) − 14 = 18 − 14 = 4 = R.H.S.

Hence, m = 6 is a solution of the given equation.