Question

Solve the following equations by trial and error methor"
i 5p + 2 = 17 (i) 3m - 14 = 4​

Answer 1

Answer: (i)3

(ii)6

Step-by-step explanation:

(i) 5p+2+17

let p=1

Substitutiong p=1,

5(1)+2=17

5+2=17

7$\neq$17

let p=3

substituting p=3,

5(3)+2=17

15+2=17

17=17

LHS=RHS

∴p=3

(ii)3m-14=4

let m=3

Substituting m=3,

3(3)-14=4

9-14=4

-5$\neq$4

let m=6

substituting m=6,

3(6)-14=4

18-14=4

4=4

LHS=RHS

∴m=6

Answer 2

Answer:

1 let p be 1

5×1 +2 is not equal to the 17

let p be 2

5×2 +2 =12

let p be 3

5×3 +2 =17

value of p is 3

2. let m be 1

3×1 -14 is not equal to 4

let m be 2

3× 2 -14 is not 4

let m be 6

3 ×6-14=18-14 =4

value of m is 6......