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Thread: College Algebra

  1. #21
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    Re: College Algebra

    I'm about to go to bed , but will be up early to finish the homework questions that are giving me trouble, I am listing the problems I need assistance with , just need help on how to start out as most I'm not sure exactly what their asking or where to start. If anyone can help with the following please post and I will be back on in the morning. Thanks so much, I am so excited about finding this web site, it has helped me soooooo much!

    * solve for x: (2/x-1)+1= 2/x^2-x

    * solve for b: 5b^2-5b+1=0

    *Let f(x)= (5/b+5)-(2/b). Find the domain and simplify

    *Simplify: (t^2-25/t^2+8t+15)/(t-5/t+9)

    *Simplify: (-27)^(1/3)

    * Arlene wishes to invest $5000. If she invests part at 7% simple interest, part at 6%, and receives a total of $332 after one year, how much does she invest at each rate?

  2. #22
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    Re: College Algebra

    MMM4444BOT.... WHERE DID YOU GET HE -2X AS THERE IS NO X IN THE NUMERATOR?

  3. #23
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    Re: College Algebra

    HAVE TO GO TO BED HOPE TO FINISH IN MORNING, AGAIN THANKS FOR ALL YOUR HELP.

  4. #24
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    Quote Originally Posted by twhitley

    MMM4444BOT.... WHERE DID YOU GET HE -2X AS THERE IS NO X IN THE NUMERATOR?
    Quote Originally Posted by mmm4444bot

    Do you know how to combine the two ratios on the righthand side ?

    Get a common denominator. Multiply the second ratio by x/x.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  5. #25
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    Quote Originally Posted by twhitley

    * solve for b: 5b^2-5b+1=0

    *Let f(x)= (5/b+5)-(2/b). Find the domain and simplify

    *Simplify: (t^2-25/t^2+8t+15)/(t-5/t+9)

    *Simplify: (-27)^(1/3)

    * Arlene wishes to invest $5000. If she invests part at 7% simple interest, part at 6%, and receives a total of $332 after one year, how much does she invest at each rate?
    Please do not post lists of exercises with no work shown. Have you learned anything about these, yet? I can't tell what kind of help you need. Also, please post each new exercise in its own thread.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  6. #26
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    Re: College Algebra

    3)solve for x: (2/x-1) +1=(2/x^2-x)

    No, I don't understand this one. I get subtracting the 2/x-1 and understand how you end up with 1= 2/x^2-x-2/x-1 and am lost from there. Yes, I am learning something and that's why I'm here is for help, not looking for answers, I want to get what I'm doing and sorry for posting all of them at once, new to this message board thing. I'll be out for awhile, but will check in later. Thanks.

  7. #27
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    Re: College Algebra

    Quote Originally Posted by twhitley
    3)solve for x: (2/x-1) +1=(2/x^2-x)
    Please use brackets PROPERLY; your equation should be typed this way:
    2/(x - 1) + 1 = 2/(x^2 - x)

    20 / 2 * 5 = 10 * 5 = 50
    20 /(2 * 5) = 20 / 10 = 2
    Get my drift?

    Hint: x^2 - x = x(x - 1) ; let's see what you can do....
    I'm just an imagination of your figment !

  8. #28
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    (2/x-1)+1= 2/x^2-x ? This is not properly typed.

    You need to put grouping symbols around denominators.

    My guess is: 2/(x - 1) + 1 = 2/(x^2 - x)


    2/(x - 1) - 2/(x^2 - x) = 1

    You should know that two ratios cannot be combined unless they share a common denominator.

    If we multiply 2/(x - 1) by x/x, then both ratios will have the same denominator:

    2/(x - 1) * x/x = 2x/(x^2 - x)

    See the denominator? It's now the same as the other one.

    2x/(x^2 - x) - 2/(x^2 - x) = 1

    Since both denominator are the same, we can combine the two ratios into a single ratio.

    (2x - 2)/(x^2 - x) = 1

    I also previously explained why we can now write the following equation:

    2x - 2 = x^2 - x

    Solve this equation for x, and make sure that you CHECK your solutions.

    Please repost your other exercises, each in their own NEW thread.

    Cheers ~ Mark

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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