Question

solve the equation 2y+4=14 ;m–10=2 by the method of trial and error​

Answer 1

Answer:

Solution 3:

(i) putting p = -3 in equation

5(-3) +2 = 17

-15+2 = 17

-13\ne17−13

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=17

putting p = -2 in equation

5\left(-2\right)+2=175(−2)+2=17

-10+2=17−10+2=17

-8\ne17−8

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=17

putting p =-1

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

-5+2=17−5+2=17

-3\ne17−3

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=17

putting 0 in the equation

5\times0\ +2\ =\ 17\5×0 +2 = 17

2\ne172

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=17

putting p =1 n the equation

5\times1+2=175×1+2=17

5+2=175+2=17

7\ne177

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=17

puttin p = 2 in the equation

5\times2\ +2\ =\ 17\5×2 +2 = 17

10+2\ =\ 17\10+2 = 17

12\ne1712

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=17

putting p = 3 in the equation

5\times3+2=175×3+2=17

15+2=1715+2=17

17=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\ 4−20

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= 4

putting m = -1

3(-1) -14 = 4

-3-14 = 4

-17\ne4−17

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=4

putting m = 0

3(0) -14=4

0-14 = 4

-14\ne4−14

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=4

putting m =1

3(1) -14 = 4

3-14 = 4

-11\ne4−11

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=4

putting m = 2

3(2) - 14 = 4

6-14=4

-8\ne4−8

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=4

putting m = 3

3(3) - 14 = 4

9-14 = 4

-5\ne4−5

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=4

putting m = 4

3(4)- 14 = 4

12-14 = 4

-2\ne4−2

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=4

putting m = 5

3(5)-14=4

15-14=4

1\ne41

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=4

putting m = 6

3(6)-14 = 4

18-14

4=4

LHS = RHS

therefore m = 6 is a solution.

Step-by-step explanation:

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