Alright, so I’m tutoring someone, and they can’t get this problem. I (apparently) can’t either. Having been in Cal II, makes me feel stupid, because the girl I’m tutoring is in intermediate Algebra.
Problem:
“When three times the smaller of 2 consecutive even intergers is added to the larger, the result is 4 more than 3 times the larger. What are the 2 numbers?”
My method:
3(2n)+2n+2=3(2n+2)+4
8n+2=6n+10
2n=8, n=4.
Numbers are 4 and 6
But, if you put those back into the original equation, they don’t work. Where did I go wrong?
The integers have to be consecutive even integers, so 6 and 7 don’t work.
You also don’t have to use 2n for your smaller number, which is the step that tripped you up. Even though the number is even, using 2n doesn’t help - if the answer were odd, you would just get n = 1.5 or something, which would be odd when multiplied by 2. Using 2n instead of n does not give you any advantage.
Just using n and n + 2 would have given you the answer of 8 , 10 easier.
I don’t see why you are going with 2n for the smaller integer. Have n represent the smaller integer, n+2 then represenets the larger. Even or odd has nothing to do with it except that we know that the larger is 2 more than the smaller.