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Math Help - Solving For X - Algebra

  1. #1
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    Solving For X - Algebra

    Alright, I kind of zoned out in class the other day, and I can't remember the method to solve for x in this equation.

    3 + 2[4x − 3(5 − x)] = 3(x − 20)

    The answer to this question is x = -3 but i can't for the life of me figure out the right steps in order to get there, been trying for hours haha.

    If anyone can explain to me the steps in detail how to show my work for this type of question it would be much appreciated!
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  2. #2
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    e^(i*pi)'s Avatar
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    Use the order of operations

    3+2[4x-15+3x] = 3x-60

    3 + 8x-30+6x = 3x-60

    Collect like terms to simplify
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  3. #3
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    Damn, that's so easy after you said that.

    Well I'm officially a big newbie at math XD, thanks a bunch, hopefully I won't have to plague you guys with anymore basic questions haha
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  4. #4
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    Arghhh... alright so using that method I've gotten stuck again, if you could tell me what i'm doing wrong it would be MUCH appreciated.

    5 + [1 + 2 (2x - 3)] = 6 (x + 5) < that is the question, now i'll show you my work below it.

    5 + [1 + 4x - 6) = 6x + 30

    5 + 20x - 30 = 6x + 30

    20x - 6x = 30 + 30 - 5

    14x = 55


    This is where I'm stuck, the answer is " x = -15 " and I'm obviously not getting that with my answer.

    Thanks!
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  5. #5
    MHF Contributor Unknown008's Avatar
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    From this:

    5 + [1 + 4x - 6] = 6x + 30

    To this is not good.

    5 + 20x - 30 = 6x + 30

    This would become:
    5 + [1 + 4x - 6] = 6x + 30

    5 + 1 + 4x - 6 = 6x + 30

    Why did you multiply?
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  6. #6
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    Hmm, it just made sense in my mind to multiply the 5 into the brackets haha, I didn't know I could just disperse the brackets and add it all together O_o

    Thanks though XD
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Don't overdo yourself Keep cool.
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  8. #8
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    Alright, last question I promise lol sorry to bug you guys with these questions!

    2x^2 - 3x - 5
    -------------
    x^5 y^2

    x

    x^4 y^3
    ----------------
    2x^2 - 7x + 5


    I have to "Perform the indicated operations and simplify"

    These big equations just destroy my mind, I need to learn the proper steps to tackle these suckers, my math skills have too many gaps >_< lol

    Thanks again!

    EDIT:: I put the question vertically as it was getting messed up horizontally, so it's the top multiplied by the bottom.
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  9. #9
    MHF Contributor Unknown008's Avatar
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    Okay, this is it then?

    \dfrac{2x^2-3x-5}{x^5y^2} \times \dfrac{x^4y^3}{2x^2 - 7x + 5}

    You can start by simplifying the denominator of the first fraction with the numerator of the second fraction, or move it like this to make it easier:

    \dfrac{x^4y^3}{x^5y^2} \times \dfrac{2x^2-3x-5}{2x^2 - 7x + 5}

    Then, factorise the second fraction.

    What do you get?
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  10. #10
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    e^(i*pi)'s Avatar
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    \dfrac{2x^2 - 3x-5}{x^5y^2} \times \dfrac{x^4y^3}{2x^2-7x+5} (double click the code to see how to write it)

    2x^2-3x-5 = (2x-5)(x+1) \text{    and    } 2x^2-7x+5 = (2x-5)(x-1)
    Last edited by e^(i*pi); November 14th 2010 at 09:01 AM. Reason: I thought fraction read -5 not +5
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  11. #11
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    Haha, I think you guys are assuming I'm more intelligent at math than I am XD

    Even seeing what you guys are doing, I don't understand or even know what rules or methods you're using that enable you to do what you do. You guys can probably see the answer by just looking at it, but for me it's just a bunch of numbers and variables being rearranged haha.
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  12. #12
    MHF Contributor Unknown008's Avatar
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    Well, let's take a simple example:

    \dfrac12 \times \dfrac35

    Do you agree this can also be written as:

    \dfrac32 \times \dfrac15

    ?

    This is because when you multiply you get:

    \dfrac{1\times3}{2 \times 5}

    And since multiplication is commutative, that it you can change the order of the numbers without altering the result.

    I used the same principle here.

    Then, since you see a quadratic expression, you should be able to factorise it, just to see if there is something which may cancel out, and in your case, there is
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