find 3 consecutive integers such that the difference of 5 times the third and 5 more than the three times the secound is 36

Answer 1

first integer = x

second integer = x+1

third integer = x+2

5(x+2) - [3(x+1)+5] = 36

5x+10 -[3x+3+5] = 36

5x+10-3x-8 =36

2x+2 = 36

2x = 34

x =17

first integer = x =17

second integer = x+1 =18

third integer = x+2=19

Answer 2

Three consecutive integers are represented by X, X+1, X+2

Five times the third is 5(X+2)

3 times the second and 5 is 3(X+1)+5

Difference 5(X+2) - (3(X+1)+5) = 36

5X+10 - (3X+3+5) = 36

5X+10-3X-8 = 36

2X+2=36

2X=34

X=17

Consecutive integers are 17, 18, 19

5*19 = 95

3*18+5=59

95-59=36