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Math Help - simple college algebra

  1. #1
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    simple college algebra

    I have gone back to school after 30 yrs and need a little help on a couple of problems. I know they are simple but for some reason I am have problems with these two. I can find the answers in the back of the text book but that does me no good. If someone could show me how to solve the it woould be greatly apprieciated. thanx. Here they are. I dont know how to show something is squared so I will just put the exp. after it.

    1) 5/x2-43/x=18 ( the first x is squared)
    2) 2=3/2x-1+-1/(2x-1)2 ( the 2x-1 is squared)
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  2. #2
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    1) \frac {5}{x^2} -\frac {43}{x}=18
    2) 2 =\frac {3}{2x-1} -\frac {1}{(2x-1)^2}

    For problem 1 multiply through by a factor of x^2. This gets 5 - 43x=18x^2 Then apply the quadratic formula (a=18, b=43, c=-5)to obtain your values for x. The second problem is similar yet you multiply through by a factor of (2x-1)^2, expand and then simplify.
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  3. #3
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    Thanx for the help
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  4. #4
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    Quote Originally Posted by dinNA89 View Post
    1) \frac {5}{x^2} -\frac {43}{x}=18

    For problem 1 multiply through by a factor of x^2. This gets 5 - 43x=18x^2 Then apply the quadratic formula (a=18, b=43, c=-5)to obtain your values for x.
    Or, you could factor. Just for fun.

    18x^2+43x-5=0

    18x^2+45x-2x-5=0

    (18x^2+45x)-1(2x+5)=0

    9x(2x+5)-1(2x+5)=0

    (9x-1)(2x+5)=0

    9x-1=0 \; \;\Longrightarrow x=\frac{1}{9}

    2x+5=0 \; \; \Longrightarrow x=\frac{-5}{2}
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  5. #5
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    I know you said just for fun it could be factored, but at this point for me that would have been anything BUT fun. I do thank you for the help though and will more than likely be needing it again.
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  6. #6
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    Quote Originally Posted by whiteowl View Post
    ... If someone could show me how to solve the it woould be greatly apprieciated. thanx. Here they are. I dont know how to show something is squared so I will just put the exp. after it.

    ...
    2) 2=3/2x-1+-1/(2x-1)2 ( the 2x-1 is squared)
    2=\frac3{2x-1}+\frac{-1}{(2x-1)^2}

    Use the substitution y = \frac1{2x-1} . Your equation becomes:

    2=3y-y^2~\iff~y^2-3y+2=0

    Now use the formula to solve quadratic equations or factor the LHS. You should get:

    y = 1 or y = 2

    Plug in these values into the substitution and solve for x:

    1 = \frac1{2x-1}~\implies~x = 1

    or

    2 = \frac1{2x-1}~\implies~x = \frac34
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