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

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

the answer

b. they are parallel and will never intersect.

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1