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Algebra II Simplifying Rational Algebraic Expressions

  1. Mar 7, 2012 #1
    Hi everyone and nice to meet you. I'm velox_xox a newbie to PF and a high school correspondence student currently taking Algebra II. Since I'm correspondence, I am basically teaching myself my subjects, which means if I don't understand something it's a big problem. I enjoy math, but I must admit it is one of my weaker subjects so any help on this and future questions would be GREATLY appreciated. Thank you!

    1. The problem statement, all variables and given/known data
    Simplify: x^2 - y^2 - 4y - 4 / x^2 - y^2 -4x + 4
    The answer (according to the back of the book) should be: x + y + 2 / x + y - 2

    2. Relevant equations

    3. The attempt at a solution
    I've tried multiple combinations to try to get the correct answer. Here is my closest one:

    x^2 - y^2 - 4y - 4 / x^2 - y^2 - 4x + 4 =
    Using the difference of the squares and distributive property to factor I got:
    (x + y)(x - y) -4(y +1) / (x + y)(x-y) -4(x-1) =
    Simplify by getting rid of like terms:
    (y + 1) / (x - 1)

    I'm pretty sure I'm missing some combination between GCF, factoring, Perfect Square Trinomials, and/or Difference of Squares. I just can't see which ones to use that will give me the right answer.
  2. jcsd
  3. Mar 7, 2012 #2


    Staff: Mentor

    Welcome to PF! You're off to a good start by including the homework template, and showing what you have tried. I think you will find PF very helpful.

    When you write expressions like the ones above on a single line, put a pair of parentheses around the entire numerator, and another pair around the entire denominator.

    For example, what you're trying to simplify should be written as (x^2 - y^2 - 4y - 4) / (x^2 - y^2 -4x + 4).

    The answer that the book shows should be written as (x + y + 2) / (x + y - 2). Otherwise it would be interpreted as x + y + (2/x) + y - 2, which is not what was intended.
    From where you started, rewrite the expression as (x^2 - (y^2 + 4y + 4))/((x^2 - 4x + 4) - y^2). Is this enough of a start?
  4. Mar 8, 2012 #3


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  5. Mar 10, 2012 #4

    First off, thank you Mark44 for your timely response. I apologize about the brackets. I'll be sure to do that from now on.

    I'm really close, but somehow now I have it so that the answer is '1'.

    Here's what I got:

    Using your hint as the starter:
    (x^2 - (y^2 + 4y + 4))/((x^2 - 4x + 4) - y^2)
    Next I factored because I didn't realize right away that in both the numerator and denominator there is a perfect square trinomial.
    (x^2 - (y+2)(y+2)) / ((x - 2)(x - 2) - y^2)
    Then, I recognized it, so I put it into the perfect square trinomial form.
    (x^2 - (y + 2)^2) / ((x - 2)^2 - y^2)
    The whole thing can be simplified by a square root, which leaves:
    (x - (y + 2)) / ((x - 2) - y))
    When I get rid of the brackets and reorder it in the standard form (of x, then y, then numericals), I got this:
    (x - y - 2) / (x - y - 2)
    And that simplifies to '1' as I said.

    Did I do something wrong with the distributive property, or am I missing something really obvious?

    Also, HallsofIvy... I assume you are talking about Mark44's hint, right? If not, it's unclear what is 'lovely.'
  6. Mar 10, 2012 #5


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    Staff: Mentor

    Good. https://www.physicsforums.com/images/icons/icon14.gif [Broken]

    http://img140.imageshack.us/img140/9826/t2622.gif [Broken]

    Think about it; try a concrete example: √(4 + 9)
    Does this equal 2 + 3 ?
    Last edited by a moderator: May 5, 2017
  7. Mar 10, 2012 #6
    Ah, I see. I was so caught up in the algebra part of it that I forgot the basic orders of operation.

    So, going from the perfect square trinomial step, this:
    (x^2 - (y + 2)^2) / ((x - 2)^2 - y^2)
    I use the second law of exponents (and square both terms in the inner brackets):
    (x^2 + y^2 + 4) / (x^2 + 4 - y^2)
    Reorder into the traditional algebra form:
    (x^2 + y^2 + 4) / (x^2 - y^2 + 4)
    :confused: Then, simplify... The numerator makes sense as:
    (x + y + 2)

    But I'm still not understanding how on the denominator the (-y^2) becomes a (+ y), and the '4' becomes a '-2'. I do understand that a square could be with a negative or positive number. Ie: '2' or '-2' squared both equal a postive '4'. I just feel like I'm bending the rules to make this work.
    Last edited: Mar 10, 2012
  8. Mar 10, 2012 #7


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    jey, please don't give out full solutions, it's against the forum rules.

    velox, you've made numerous algebraic errors in hopes of reaching the right answer. Remember that you need to follow the rules you learnt from Algebra I exactly. There are no shortcuts, and if you're finding it hard to reach the answer in some of your questions, you should still especially hard here to avoid to falling into the trap of making some (wrong) simplifications that seem to get you closer to the answer.

    For example,
    No, you cannot just take the square root like that.

    [tex]\sqrt{a^2+b^2}\neq a+b[/tex]
    Always keep this rule in mind. By the way, it's true because if we go backwards and square both sides then we'll get

    [tex]LHS= \left(\sqrt{a^2+b^2}\right)^2[/tex]


    [tex]RHS = (a+b)^2[/tex]


    [tex]\neq LHS[/tex]

    Now for the second example:
    You factorized correctly, but then went to expand back out again (which is just reversing the process of what you already did, hence getting you nowhere). The problem here is even worse however, you expanded incorrectly.



    Another example,
    No, it doesn't make sense. Remember, only use the rules you learnt from Algebra I. There's a reason they get you to solve so many questions, so the rules will be imprinted into your head so that when you encounter harder problems, you can reproduce those rules more easily.

    Exactly :wink:

    Go back to the expression


    Do you see what can make things simpler? Factorize maybe? :smile:
    Last edited: Mar 11, 2012
  9. Mar 10, 2012 #8


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    Staff: Mentor

    If there is the slightest doubt in your mind that a step is valid, try some numeric values (as I showed with the √ ) to validate it. (Not 100% foolproof, but better than keeping your fingers crossed and hoping for the best. Without a teacher there to spot errors early, you have to exercise that vigilance yourself!)

    https://www.physicsforums.com/images/icons/icon2.gif [Broken] A useful and oft-forgotten expansion to apply now: (x2 - a2) = (x-a)·(x+a)
    Last edited by a moderator: May 5, 2017
  10. Mar 11, 2012 #9
    Aaah, that's part of the problem. My Algebra I teacher (not me), didn't make it through all of the Algebra I class in the school year. So I don't think I learned everything there was to learn. Plus, it's been a while since I had that class. I'll try not to bend the rules.

    So... is this on the right path?

    Where I left off:
    {x^2 - (y + 2)^2} / {(x - 2)^2 - y^2}
    Then, the hint Mentallic gave moved the inner parenthesis:
    {x^2 - (y + 2)^2} / {x^2 - (y - 2)^2}
    Which as NascentOxygen pointed out is the difference of the squares (?), so:
    {x + (y + 2)}{x - (y + 2)} / {x + (y - 2)}{x - (y - 2)}
    Is it okay to drop the inner parentheses?
    (x + y + 2)(x - y + 2) / (x + y - 2)(x - y - 2)

    If so, then I'm really close to the answer. But I personally have looked at it and don't see a method of further simplification that is correct.

    By the way, if anyone notices one point algebraically or even basic arithmetic that I seem to get tripped up on. Let me know. I'll go back to those sections and study up.
  11. Mar 11, 2012 #10


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    That's good to hear :smile:
    And you should definitely follow NascentOxygen's advice of testing random values between each step you take in case you're unsure if you're breaking the rules.
    I know I used to do it when I was rusty.

    Oh no sorry! I misread the denominator as being x^2-y^2-4y+4 as opposed to x^2-y^2-4x+4. Stick to (x - 2)^2 - y^2!
    As a note, you can't just move the inner parenthesis like that. If you expand out (x - 2)^2 - y^2 and then x^2 - (y - 2)^2 you'll notice that they give different answers.
    And besides, x^2 - (y - 2)^2 doesn't even expand to give x^2-y^2-4y+4 so that was a real screw up on my part. So don't feel hard on yourself if you make some mistakes, because we all do :biggrin:

    You must treat it as being any other parenthesis.
    How would you handle x-(y-z) ?

    Yes you are really close! Once you treat each parenthesis correctly, you'll see a nice simplification that you can do.

    p.s. I'll edit the mistake in my last post.
  12. Mar 11, 2012 #11
    I thought that wasn't quite right. I'm glad you caught it, though! Phew!

    I would handle it by carrying out the distributive property. It is basically:
    x -1(y - z)
    So, distributive property:
    x -1y +1z
    x - y + z


    So, starting with the correct difference of the squares...
    {x + (y+2)}{x - (y +2)} / {(x-2) + y}{(x - 2) - y}
    Distributive property of 1 (to eliminate parentheses):
    {x + 1y +(1)2}{x - 1y -1(2)} / {1x +1(-2) + y}{1x + 1(-2) - y}
    (x + y + 2)(x - y - 2) / (x + y - 2)(x - y - 2)
    (x + y + 2) / (x + y - 2)

    Is this correct? Not only the answer but the process (a.k.a. did it follow the algebraic rules)?

    P.S. I know most people skip the whole "1y" thing, but I did that to show how I arrived at my answer for the sake of making sure I didn't luck into the right answer!
  13. Mar 11, 2012 #12


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    Yes, that's correct :smile:

    Just a little tip though, placing a 1 outside the parenthesis before expanding is unnecessary, so if you can do it without, definitely go for it.

    For example, you can just as well do

    [tex]x-(y-z) = x -y -(-z) = x-y+z[/tex]

    But once you get more used to it, you'll be skipping the middle part and just going straight to the last expression.
  14. Mar 11, 2012 #13
    sorry, won't happen again.
  15. Mar 12, 2012 #14
    Whoo! I finally got it!

    So a very hearty thanks to Mark44, NascentOxygen, Mentallic, and jey1234 (apparently, I missed your original post, so no harm no foul). Thank you everyone for your help!

    Unfortunately (or fortunately, depending on which way you look at it), this probably won't be the last you hear from me in this subforum! I'm sure I'll be back with new algebra puzzles for my mind. :)

    Until next time...
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