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Math Help - Rational equations problem help?

  1. #1
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    Rational equations problem help?

    I am having difficulty solving these equations

    (7A/3A+3)-(5/4A-4)=(3A)/(2A+2)

    (2Q/(2Q+3)-(2Q/2Q-3)=1

    and

    (12/t)+t-8=0
    Can anyone help me solve these and show me how to as I am stuck on rational equations I am not good at it.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by homeylova223 View Post
    I am having difficulty solving these equations

    (7A/3A+3)-(5/4A-4)=(3A)/(2A+2)

    (2Q/(2Q+3)-(2Q/2Q-3)=1

    and

    (12/t)+t-8=0
    Can anyone help me solve these and show me how to as I am stuck on rational equations I am not good at it.

    you have ...
     \displaystyle \frac {12}{t} + t - 8 = 0

    so multiply by "t" to loose fraction

     \displaystyle 12 + t^2 -8t = 0

    and that's

    t^2-8t+12 = 0

    same as that you have

     ax^2+bx+c = 0

    solutions would be

     \displaystyle x_{1,2} = \frac {-b \pm \sqrt{b^2-4ac}}{2a}

    and for first two you can do the same thing ... with what you should multiply (the whole ) equation to lose fractions ? (if fractions are thing that you have problems with)
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  3. #3
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    Right now I am having trouble figuring out (7A/3A+3)-(5/4A-4)=(3A)/(2A+2)
    I think you can use 24a^2+12a^2-12a-24 as a common factor but then I get stuck when trying to solve it. This is what I usually get when I am done
    56a^3+4a+30a^2+60a+3=36a^3-36a but I am not sure if this is right.
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  4. #4
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    Quote Originally Posted by homeylova223 View Post
    Right now I am having trouble figuring out (7A/3A+3)-(5/4A-4)=(3A)/(2A+2)
    I think you can use 24a^2+12a^2-12a-24 as a common factor but then I get stuck when trying to solve it. This is what I usually get when I am done
    56a^3+4a+30a^2+60a+3=36a^3-36a but I am not sure if this is right.
    \displaystyle \frac{7a}{3(a+1)} - \frac{5}{4(a-1)} = \frac{3a}{2(a+1)}

    common denominator is 12(a+1)(a-1) ...

    \displaystyle \frac{7a \cdot 4(a-1)}{12(a+1)(a-1)} - \frac{5 \cdot 3(a+1)}{12(a+1)(a-1)} = \frac{3a \cdot 6(a-1)}{12(a+1)(a-1)}

    try it again ...
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  5. #5
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    After doing that I get
    28A/12(A+1)-15/12(a-1)= 18A/12(a+1)

    Now from the back of the book I know my final answer are (-1/2 and 3). I am just no sure how to get there.
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  6. #6
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    Skeeter's point is that after you have the same denominator for every fraction, you can multiply through by that denominator and it disappears!

    From \displaystyle \frac{7a \cdot 4(a-1)}{12(a+1)(a-1)} - \frac{5 \cdot 3(a+1)}{12(a+1)(a-1)} = \frac{3a \cdot 6(a-1)}{12(a+1)(a-1)}, multiply each part by 12(a+1)(a-1) and you have 7a(4)(a-1)- 5(3)(a+1)= 3a which is the same as 28a^2- 28a- 15a- 15= 3a.
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  7. #7
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    Then I get 28a^2-46a-15
    I am not sure how to factor this to get to -1/2 and 3 unless I have to use quadratic formula. Maybe I have to multiply the 3a(6)(a-1).
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  8. #8
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    Quote Originally Posted by homeylova223 View Post
    Then I get 28a^2-46a-15 ... no.
    I am not sure how to factor this to get to -1/2 and 3 unless I have to use quadratic formula. Maybe I have to multiply the 3a(6)(a-1).
    correction ...

    10a^2 - 25a - 15 = 0

    5(2a^2 - 5a - 3) = 0

    5(2a + 1)(a - 3) = 0
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