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Math Help - Little Indices Problem

  1. #1
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    Little Indices Problem

    Hello again guys,

    I have five problems to solve but found myself stuck by this question other thought to be very simple but I am unfortunately not. Could you guys explain it how to solve this problem? I don't know how to make the base number the same because both a and b is variable and not number. Thank you very much for your help.

    Solve the following simultaneous equations:

    75 = ab^2, 275 = ab^3

    My first time using the Latex software, so please forgive me if I've made a mistake.
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  2. #2
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    Quote Originally Posted by Lites View Post
    Hello again guys,

    I have five problems to solve but found myself stuck by this question other thought to be very simple but I am unfortunately not. Could you guys explain it how to solve this problem? I don't know how to make the base number the same because both a and b is variable and not number. Thank you very much for your help.



    Solve the following simultaneous equations:

    75 = ab^2, 275 = ab^3

    My first time using the Latex software, so please forgive me if I've made a mistake.
    b\left(ab^2\right)=ab^3\ \Rightarrow\ b=\frac{275}{75}=\frac{11}{3}

    Then a=\frac{75}{b^2}
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    b\left(ab^2\right)=ab^3\ \Rightarrow\ b=\frac{275}{75}=\frac{11}{3}

    Then a=\frac{75}{b^2}
    Thank you for your time Archie Meade, you've always come to help. But when I check it again, the key of the book says Do you know how do they get those answers?

    a = 3, b = 5?

    I'm sorry if I don't understand you fully.
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  4. #4
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    What I would do is "divide one equation by the other":
    \frac{275}{75}= \frac{ab^3}{ab^2}= b

    \frac{275}{75}= \frac{11(25)}{3(25)}= \frac{11}{3} as Archie Mead said.

    Once you know b, replace b in either of the equations by that and solve for a.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    What I would do is "divide one equation by the other":
    \frac{275}{75}= \frac{ab^3}{ab^2}= b

    \frac{275}{75}= \frac{11(25)}{3(25)}= \frac{11}{3} as Archie Mead said.

    Once you know b, replace b in either of the equations by that and solve for a.
    I just want to clarify several things,

    What would you get after dividing both equation with each other? To know the B right? And the B is \frac{11}{3}
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  6. #6
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    Quote Originally Posted by Lites View Post
    Thank you for your time Archie Meade, you've always come to help. But when I check it again, the key of the book says Do you know how do they get those answers?

    a = 3, b = 5?

    I'm sorry if I don't understand you fully.
    These answers can't be the solutions to the system of equations
    75 = ab^2
    275 = ab^3
    ... because if you plug in a and b into the 2nd equation you will get 375 on the right side, not 275. So maybe the original problem should have been
    75 = ab^2
    375 = ab^3
    ... and there was a typo?
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  7. #7
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    There was a typo indeed, I'm so sorry, it should be 375. Thank you for your kind and patient help guys.
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