PEMDAS

On my assesment it had a question that said x+3/3, when x=3. I thought you do 3/3 first so did the teacher, but when he looked at the answer key it showed that you were suppose to do 3+3 then divide by 3. And also on my test booklet, it doesn't have 4 which is the answer if you do it with PEMDAS. Instead they have 2, the answer you get when you add first... :?: Isn't that weird? The same thing happened on my state test. On one of the sample answers the state test gives a question just like that and when my teacher and the student found out the answer on the sample questions, the answer only works if you add the two numbers then divide. Also the state test doesn't have the answer I got, when I did it with PEMDAS. Instead it had the answer if you added first...Who's right and who's wrong and what is the correct way to do this? :|

x+3/3 means \(\displaystyle x+\frac{3}{3}\).
(x+3)/3 means \(\displaystyle \frac{x+3}{3}\).
Following PEMDAS (and really there is no other way), if x = 3 the first one becomes 4 and the second one becomes 2.

Loren said:

x+3/3 means \(\displaystyle x+\frac{3}{3}\).
(x+3)/3 means \(\displaystyle \frac{x+3}{3}\).
Following PEMDAS (and really there is no other way), if x = 3 the first one becomes 4 and the second one becomes 2.

Click to expand...

Is there an a time where you don't follow PEMDAS?

Do you think that the test didn't want me to follow PEMDAS? On the sample question the answer key showed the answer was 2, and there was a similar question on the test and I didn't use PEMDAS, since the sample question didn't...I'm just hoping I got it right. :?

Petenerd said:

Loren said:

x+3/3 means \(\displaystyle x+\frac{3}{3}\).
(x+3)/3 means \(\displaystyle \frac{x+3}{3}\).
Following PEMDAS (and really there is no other way), if x = 3 the first one becomes 4 and the second one becomes 2.

Click to expand...

Is there an a time where you don't follow PEMDAS?

Do you think that the test didn't want me to follow PEMDAS? On the sample question the answer key showed the answer was 2, and there was a similar question on the test and I didn't use PEMDAS, since the sample question didn't...I'm just hoping I got it right. :?

Click to expand...

I think the question looked like this \(\displaystyle \frac{x+3}{3}\), instead of \(\displaystyle x+\frac{3}{3}\). And the same thing with the other questions.

If the question meant \(\displaystyle \frac{x+3}{3}\), do I do add first? And then divide?

Petenerd said:

If the question meant \(\displaystyle \frac{x+3}{3}\), do I do add first? And then divide?

Click to expand...

Yes - the long division line implies parentheses around the numerator and the denominator separately. In ther words you can think:

\(\displaystyle \frac{x+3}{3} \, = \, \frac{(x+3)}{(3)} \,\)

or you can do the following:

\(\displaystyle \frac{x+3}{3} \, = \, \frac{x}{3} \, + \, \frac{3}{3}\)

The bar in a fraction is a grouping symbol. It works the same as if parenthesis were there. For instance---
\(\displaystyle \frac{a+b}{c-d}\) means \(\displaystyle (a+b)/(c-d)\).