Find three consecutive integers such that three times the sum of the first two is eight more than 5 times the third.

Consecutive integers are numbers one after the other on the number one (ie. 1,2,3,4,5 or 203,204,205).

times= multiplication
sum= addition
more than= addition
is= equal sign

x= first integer
x+2= second integer
x+3= third integer

3(x + (x + 2))= 5(x + 3) + 8
add inside parentheses

3(2x + 2)= 5(x + 3) + 8
multiply 3 and 5 by their parentheses

(3*2x) + (3*2)= (5*x) + (5*3) + 8
multiply in parentheses

6x + 6= 5x + 15 + 8

6x + 6= 5x + 23
subtract 5x from both sides

x + 6= 23
subtract 6 from both sides

x= 17 first integer

x + 1= 17 + 1= 18 second integer

x + 2= 17 + 2= 19 third integer

CHECK:
3(x + (x + 2))= 5(x + 3) + 8
3(17 + (17 + 2))= 5(17 + 3) + 8
3(17 + 19)= 5(20) + 8
3(36)= 100 + 8
108= 108

ANSWER: The three consecutive integers are 17, 18 and 19.

Hope this helps! :)

The first answer is correct