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Making X the subject. Know nothing need help

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello. This is my first post, hoping its in the right section. Now when it comes to physics I dont really have a clue, biology is my strong point. But still need to know this. When making X the subject what do I need to learn? I know the basics such as y = X - 3 is X = y + 3 ect. But need to know when to divide both sides of the equation by another thing in in the equation. Or what do I do when there is a squared or brackets involved. Sorry for my stupidity. Im trying. Any help appreciated. Thanks for reading


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 8, 2012 #2

    Doc Al

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    Sounds like you need to brush up on basic algebra.
     
  4. Sep 8, 2012 #3
    Well that's why I'm here. I just asked what steps are involved with something like something like a=(x-u)/t, or a=-3+x². Not to be told what I already knew
     
  5. Sep 8, 2012 #4

    Doc Al

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    We are not mind readers. Unless you present a specific question and show your work and where you got stuck, how are we to know how to help you?

    Since you already know that it's algebra you need help with, what steps have you taken to brush up on your algebra?

    As far as your specific examples, why not tell us what you think needs to be done in each case to isolate the x? Hint: The basic idea is to always do the same thing to each side of the equation. (See: Golden Rule of Algebra)
     
    Last edited: Sep 8, 2012
  6. Sep 8, 2012 #5

    Sorry mate, I snapped a bit there (long day). Ive been looking through my old math DVD's which were to suppose to offer an "in depth" review of the steps but turns out to be whistle stop tour of confusion. So for the equation a=(x-u)/t would I be able to times both side by t and use that to get at=x-u? Because I can happily work that out to be x=at+u. Is that right at all?
     
  7. Sep 8, 2012 #6

    Doc Al

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    Perfectly right.
     
  8. Sep 8, 2012 #7
    Well then i'm making progress. So thank you. Now to the next example a=-3+x². I'm confused as to how to remove the ² from the x². This is what's really confusing me (as said I'm not good at physics, really just biology) how would i do that?
     
  9. Sep 8, 2012 #8
    This has less to do with physics than with math. What do you know about square roots?
     
  10. Sep 8, 2012 #9

    Doc Al

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    Start by getting the x2 off by itself. How would you do that? Once you do that, you can take the square root of both sides.
     
  11. Sep 8, 2012 #10

    HallsofIvy

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    To "make x the subject" (in American English, "solve for x"), you have to "undo" whatever is done to x. And you "undo" by "doing the opposite", you are correct that x+ 3= y gives you x= y- 3 when you "make x the subject". x is NOT the subject to begin with because 3 is added to x. The opposite of "add 3" is "subtract 3" so what you really do is, starting with x+ 3= y, subtract 3 from both sides, x+ 3- 3= y- 3 or, since 3- 3= 0 (which is why 'subtract 3' is the opposite of 'add 3') we have x= y- 3.

    Now, if the problem is 5x= y, x is not "the subject" because it is multiplied by 5. Now, what do you think the "opposite" of "multiply by 5" is? I hope you immediately answered "divide by 5"! Dividing both sides by 5, (5x)/5= x= y/5. Similarly, if the equation were x/7= y, you should immediately think "multiply both sides by 7" because that is the opposite of x/7: 7(x/7)= x= 7y.

    Where we have combinations, we can use the same idea but be careful. To "undo" a succession of operations on x, we have to do the opposite things in the opposite order. For example, if the equation were y= 5x- 7 and you were asked NOT to make x the subject but just to evaluate for x= 3, say. You would replace x by 3 and have y= 5(3)- 7. Of course, by the "order of operations", you would first multiply 5 by 3, to get 15, then subtract 7 to get y= 8. To solve for x, we would do the opposite of "multiply by 5", which is "divide by 5", and the opposite of "subtract 7", which is "add 7"- and we do it in the opposite order: first add 7, then divide by 5: from 5x- 7= y, adding 7 to both sides gives 5x= y+ 7 and then dividing both sides by 5, x=(y+7)/5.

    This idea of "in the opposite order" is not just mathematics- that is true whenever we try to "undo" a complex procedure. What do I need to do if I want to get in my car and drive somewhere? Well, first, I have to unlock the car door, then open the door, get in, close the door, fasten the seat belt, put the key in the ignition, and start the engine". When I arrive wherever I am going, I have to "undo" that. Obviously, the opposite of "unlock the car door" is "lock the car door" but it would make no sense to do that first, before I had even got out! No, I would "stop the engine" first, remove the key from the ignition, unfasten the seat belt, open the car door, get out, close the car door, and lock it. In each case, I have done exactly the opposite operation, and in the opposite order.
     
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