Question

find three consecutive integers such that the second is eight less than twice the third. only an algebraic solution will be accepted.

Answer 1

first integer n

second integer n+1

third integer n+2

the equation given is

(n+1)=2(n+2)-8

n+1=2n+4-8

we need to clear n

1-4+8=2n-n

n=5

the three numbers are

5, 6,7