x^3 = 27
One student solved it in the following way:
1^3 = 1
2^3 = 8
3^3 = 27
ans = 3
There was a meeting for the teachers to discuss grading standards, and I brought up my student's answer. I got a number of responses:
"That's not algebra." "The student obviously doesn't understand the problem."
And so forth. Being somewhat of a punk, I asked if any of them had, in their years of teaching, actually taught their students to recognize when an answer is "algebra" or how a right answer of 3 differs from a wrong answer of 3. It wasn't a pleasant discussion. This was a university with a prestigious math department, and I was just some guy off the street. I chatted with my students the next day. None of them had ever been told what it means to "show your work," or any of the other important trappings of school math. They either did it or they didn't.
Thinking about it more, I would have sidestepped the issue completely, by changing the problem to:
x^3 = 26
Give me "x + (x + 1) + (x + 2) = 69" and I wouldn't know where to start. I could probably turn that word problem into that equation, but I would have no idea how to actually solve the equation algebraically.
But given the word problem alone I knew to divide by 3 and add the neighboring numbers. Now, actually dividing by 3 I'd need to use a calculator or spend a lot of time with a pencil and paper, but at least I knew the steps to solve it.
I'm awful at rote math and mental math, which is why I'm thankful I can make the computer do it for me.
"Tricks" work for simple cases. Algebra works for all cases.
If instead we made it "2 consecutive numbers and a third number that was five higher than the second number add up to 70" (i.e., 21 + 22 + 27), that's easily solved using 'tricks'. Really, the trick mention (which is what I did in my head, too) is just a rephrasing of the algebra. That is, I would subtract 4 off of the thing that is 5 higher (so that it's now 1 higher; the problem is now the same, get 3 consecutive integers), subtract 4 off of the number I'm trying to get (so 66), so the problem is now 3 consecutive integers that equal 66, and solve the same way (66/3 = 22, so -> 21, 22, 23), and then just add the four back in to the highest (21, 22, 27).
You memorize a few rules about what things you're allowed to do, then you apply the rules to simplify the problem in front of you. Sometimes, the hard part is that you don't know one of the rules you need to know. Other times, it's figuring out which of the rules you need to know that you should use. By practicing lots of problems, you get a good intuition for which direction to go, but it's not uncommon with more difficult problems to take the wrong way and end up confused.
First, we're allowed to drop the parentheses with addition (the "associative" rule for addition), so
x + x + 1 + x + 2 = 69
Now we put the similar terms together:
x + x + x + 1 + 2 = 69
How many Xs do we have? Three. Another way of writing that is 3x. So replace that part. Also, 1+2=3, so we'll replace that as well. "Apply rule and replace" is pretty much the most fundamental mathematical operation.
3x + 3 = 69
Let's get rid of the 3 by subtracting it from both sides to keep the equation balanced. On the left side the 3 cancels out (that's why we did this). On the right, we get 69-3, which is 66.
3x = 66
At this point, we just divide by 3 and simplify.
x = 66/3 x = 22
Our numbers are x (22), x+1 (23) and x+2 (24) according to how we listed them in the original problem.
