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Math Help - Trying to solve an equation, but a bit confused with changing of signs

  1. #1
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    Trying to solve an equation, but a bit confused with changing of signs

    It's been a loooong time since I've worked on math and I am stuck trying to solve an equation from my algebra class.

    Directions read, "Solve the equation:"

    t / t - 10 = t + 10 / 15

    I determined the LCD to be 15(t-10).

    Clearing fractions by multiplying both sides of the equation by the LCD, I get:

    15(t-10) {t / t - 10} = 15(t-10) {t+10 / 15}, which equals 15t = t^2 - 100.

    This is the part where I get confused, regarding changing of signs as a result of writing the equation in standard form. I got this far because I already know the answer:

    So if 15t = t^2 - 100, to rewrite in standard form I am instructed to translate the equation to:

    t^2 - 15t - 100 = 0. Where does the negative sign between t^2 & 15t come into play?

    Factored equation is (t+5) (t-20) = 0.

    On a side note, is there a way to write fractions or degrees when posting other than using ' / ' or ^ ?
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  2. #2
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    Re: Trying to solve an equation, but a bit confused with changing of signs

    Quote Originally Posted by Odonsky View Post
    It's been a loooong time since I've worked on math and I am stuck trying to solve an equation from my algebra class.

    Directions read, "Solve the equation:"

    t / t - 10 = t + 10 / 15
    First, you mean "t/(t- 10)= (t+ 10)/15" don't you? What you wrote was "1- 10= t+ (2/3)" because t/t= 1 and 10/15= 2/3.


    I determined the LCD to be 15(t-10).

    Clearing fractions by multiplying both sides of the equation by the LCD, I get:

    15(t-10) {t / t - 10} = 15(t-10) {t+10 / 15}, which equals 15t = t^2 - 100. [/quote]
    Yes, this is correct.

    This is the part where I get confused, regarding changing of signs as a result of writing the equation in standard form. I got this far because I already know the answer:

    So if 15t = t^2 - 100, to rewrite in standard form I am instructed to translate the equation to:
    You want to get "0" on the right side so subtract 15t from both sides:
    16t- 15t= t^2- 100- 15t which is the same as 0= t^2- 15t- 100. Of course, if a= b, then b= a so that is the same as
    t^2- 15t- 100= 0.

    t^2 - 15t - 100 = 0. Where does the negative sign between t^2 & 15t come into play?

    Factored equation is (t+5) (t-20) = 0.

    On a side note, is there a way to write fractions or degrees when posting other than using ' / ' or ^ ?
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  3. #3
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    Re: Trying to solve an equation, but a bit confused with changing of signs

    Quote Originally Posted by HallsofIvy View Post
    First, you mean "t/(t- 10)= (t+ 10)/15" don't you? What you wrote was "1- 10= t+ (2/3)" because t/t= 1 and 10/15= 2/3.
    Yes, you are correct.

    Quote Originally Posted by HallsofIvy View Post
    You want to get "0" on the right side so subtract 15t from both sides:
    16t- 15t= t^2- 100- 15t which is the same as 0= t^2- 15t- 100. Of course, if a= b, then b= a so that is the same as
    t^2- 15t- 100= 0.
    Ah, I see it now. I wasn't subtracting the 15t originally from on the left side of the equation. Makes perfect sense now.

    Thank you very much!
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  4. #4
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    Re: Trying to solve an equation, but a bit confused with changing of signs

    I wasn't subtracting the 15t originally from on the left side of the equation. Makes perfect sense now.

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