Hi Rob,
Set the first integer equal to n. Recall that with consecutive odd integers, the next one is always n+2, then the following one is always n+4. Same goes with consecutive even integers. Therefore:
n=1st odd integer
n+2=2nd odd integer
n+4=3rd odd integer
Now, you were given that four times the second decreased by the third is the same as two more than the first. A mouthful, but we can translate algebraically. First, four times the second:
4(n+2)
Now, decreased by the third, recall that the third integer is n+4, and decreased by implies subtraction, so:
4(n+2)-(n+4)
Finally, equal to two more than the first; more than implies addition, which means
4(n+2)-(n+4)=n+2
Let's find n.
Distributing:
4n+8-n-4=n+2
Combine like terms:
3n+4=n+2
Get n on same side:
2n+4=2
Isolate n:
2n= -2
Divide:
n=-1=first odd integer
To get the next two, add 2 to -1, then add two more to that. Integers are:
-1,1,3
You can check this by substituting back into the initial equation we created. For problems like these, English-algebra translation is half the battle. I hope this helps.
