Question

Find three consecutive integers such that the sum of the first and twice the second is 110 minus three times the third

Answer 1

First off, consecutive integer and  can be written as n, n+1 and n+2

(try putting a number in for n if it does not make sense yet)

The sum of the first and twice the second would be n+2(n+1)

So, our equation becomes

n+2(n+1)=110-3(n+2)

n+2n+2=110-3n-6

3n+2=104

6n=102

n=17

so, our three term are 17,18 and 19