# 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! :)

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