Question

Find the solution of the equation 4 +
a
3
= 5 by the trial-and-error
method

Answer 1

Answer:

Find the solution of the equation 4 +

a

3

= 5 by the trial-and-error

method

Answer 2

Solution 3:

(i) putting p = -3 in equation

5(-3) +2 = 17

-15+2 = 17

-13\ne17

putting p = -2 in equation

5\left(-2\right)+2=17

-10+2=17

-8\ne17

putting p =-1

5\times\left(-1\right)+2\ =\ 17\

-5+2=17

-3\ne17

putting 0 in the equation

5\times0\ +2\ =\ 17\

2\ne17

putting p =1 n the equation

5\times1+2=17

5+2=17

7\ne17

puttin p = 2 in the equation

5\times2\ +2\ =\ 17\

10+2\ =\ 17\

12\ne17

putting p = 3 in the equation

5\times3+2=17

15+2=17

17=17\

As LHS=RHS

p=3 is the solution

(ii) 3m-14 = 4

putting m = -2

3(-2) -14 = 4

-6-14=4

-20\ \ne\ 4

putting m = -1

3(-1) -14 = 4

-3-14 = 4

-17\ne4

putting m = 0

3(0) -14=4

0-14 = 4

-14\ne4

putting m =1

3(1) -14 = 4

3-14 = 4

-11\ne4

putting m = 2

3(2) - 14 = 4

6-14=4

-8\ne4

putting m = 3

3(3) - 14 = 4

9-14 = 4

-5\ne4

putting m = 4

3(4)- 14 = 4

12-14 = 4

-2\ne4

putting m = 5

3(5)-14=4

15-14=4

1\ne4

putting m = 6

3(6)-14 = 4

18-14

4=4

LHS = RHS

therefore m = 6 is a solution.