4

In an algebra review book, one exercise asked to solve for $x$: $$a(x+2)=\pi-cy$$ I arrived at the following: $$x=\frac{\pi-cy}{a}-2$$ The book stated the correct answer is: $$x=\frac{\pi-cy-2a}{a}$$ I see that technically I ignored the PEMDAS guideline in my answer, however, my solution doesn't seem very far from the book's answer. I'm rusty on my algebra and am curious as to whether or not my answer is acceptable or incorrect.

Jyrki Lahtonen
  • 133,153
measureallthethings
  • 153
  • 1
  • 7
  • 4
    Both your answer and the book's are exactly the same thing. – DonAntonio Oct 03 '13 at 16:40
  • 4
    Sadly, all of the discussion in the answers regarding PEMDAS is precisely the kind of pedantic, meaningless nonsense that turns students off of mathematics in the first place. – Emily Oct 03 '13 at 17:09
  • @Arkamis: agreed. My initial answer ignored it completely, but I decided it was worth looking up and verifying that it doesn't change what I said. – abiessu Oct 03 '13 at 17:36

4 Answers4

4

\begin{align*} a(x+2)&=\pi-cy \\ \Rightarrow ax&=\pi-cy-2a \\ \Rightarrow x&=\frac{\pi-cy-2a}{a} \\ &=\frac{\pi-cy}{a}-\frac{2a}{a} \\ &=\frac{\pi-cy}{a}-2. \end{align*}

Your solutions are equivalent.

J. W. Perry
  • 5,447
  • 3
  • 22
  • 28
2

Your answer is the correct value for $x$. However, you may be required to have the answer be in a particular form, e.g., a single fraction as in the book. If you can show that one is equal to the other somewhere near the end of your logic, you should be fine, just make sure that both forms are clear to whomever is reading what you have written, and that it is clear why one is equal to the other.

In general, you can solve any problem in any way that you want (unless specifically directed to use a given method), as long as your steps are clear and definitely demonstrate your result. Then, a little bit of extra writing at the end to change your answer into the same form as required by the teacher won't hurt and will help improve your recognition skills in seeing how various forms of an equation are equal.

Since you have noted that "technically I ignored the PEMDAS guideline" here is my analysis of what the difference entails:

In my opinion, the following two transformations follow PEMDAS equally, although the first would be considered "correct" as the parenthesis is dealt with directly and the second would not since division is applied before expanding due to parenthesis:

$$a(x+2)=\pi-cy=ax+2a\implies ax=\pi-cy-2a\tag{1}$$

$$a(x+2)=\pi-cy\implies x+2={\pi-cy\over a}\tag{2}$$

Again, in my opinion, dividing both sides by $a$ takes care of the parenthesis directly, so you may be able to argue that you have followed the guideline, but the transformation in $(1)$ is the more correct method if PEMDAS is required.

abiessu
  • 8,115
0

though both your answer and the book's answer are not very different but when you are told to solve for x you must use the PEMDAS guideline. but if you can show that your ans and the books answer is identical then it will be ok ie you have to do one more step!

User8976
  • 12,637
  • 9
  • 42
  • 107
0

Your answer is correct. The book's answer is merely another way of writing the same thing. Which expression is to be preferred would depend on what use (if any) is to be made of this answer in subsequent computations. If there are no subsequent computations, then I would prefer your answer over the book's, but I regard this as a personal preference, not a matter of "right" or "wrong".

Andreas Blass
  • 71,833