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    Basic Algebra Question

    Hi All,

    I've been touching up on my algebra in preparation for university and have found a knowledge gap due to me never learning a concept.

    Could someone please explain to me why -125/-8 became positive? Image below



    Thanks in advance
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    Re: Basic Algebra Question

    factor out -1 from both -125 and -8 and cancel it out
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    Re: Basic Algebra Question

    Quote Originally Posted by Katsumi View Post
    Hi All,

    I've been touching up on my algebra in preparation for university and have found a knowledge gap due to me never learning a concept.

    Could someone please explain to me why -125/-8 became positive? Image below



    Thanks in advance
    Firstly, When you had -125/-8 on LHS the minus and minus cancel out to become positive. Its basically -1/-1 multiplied by 125/8 and when you have two of the same numbers on the numerator and the denominator they yield to = 1.

    Now you have 125/8=1/r^3
    Multiply 8 to the other side
    ∴ 125=8/r^3
    Multiply r^3 to both sides
    ∴125*r^3=8
    Divide 125 to both sides
    ∴ r^3=8/125
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    Re: Basic Algebra Question

    I failed so many tests in high school because of that. If i'm thinking in the right way below then my problem was believing that 1/x^y automatically converted into just x^y without any further simplification needed.

    So fundamentally when you have r^-3 = -125/-8 you complete the following process

    1. As r has a negative coefficient you first convert it into a fraction so that it can become positive
    -1/r^-3 = -125/-8

    2. As both sides of the equation are negative - the negative exponents cancel out and the equation stays balanced
    1/r^3 = 125/8

    3. You then need to separate r^3 from the 1 which it is dividing. You do this by multiplying across the 8.
    8/r^3 = 125

    4. You then need to isolate variable r. As the coefficients 8 and 125 are on different sides of the equation this can be done in 2 operations.

    Operation One
    8 = 125 * r^3

    Operation Two
    r^3 = 8/125

    5. Now that variable r has been isolated we can find the value by taking the cube root of both sides of the equation.

    r = 2/5

    Is this the correct way of thinking or is there a concept behind it.
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    Re: Basic Algebra Question

    Quote Originally Posted by Katsumi View Post
    I failed so many tests in high school because of that. If i'm thinking in the right way below then my problem was believing that 1/x^y automatically converted into just x^y without any further simplification needed.

    So fundamentally when you have r^-3 = -125/-8 you complete the following process

    1. As r has a negative coefficient you first convert it into a fraction so that it can become positive
    -1/r^-3 = -125/-8

    2. As both sides of the equation are negative - the negative exponents cancel out and the equation stays balanced
    1/r^3 = 125/8


    3. You then need to separate r^3 from the 1 which it is dividing. You do this by multiplying across the 8.
    8/r^3 = 125

    4. You then need to isolate variable r. As the coefficients 8 and 125 are on different sides of the equation this can be done in 2 operations.

    Operation One
    8 = 125 * r^3

    Operation Two
    r^3 = 8/125

    5. Now that variable r has been isolated we can find the value by taking the cube root of both sides of the equation.

    r = 2/5

    Is this the correct way of thinking or is there a concept behind it.
    The negative in the indices cannot be cancelled. So it is -1/r^-3=125/8 and its only RHS where the negative 'goes away'
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    Re: Basic Algebra Question

    Quote Originally Posted by Rathin View Post
    The negative in the indices cannot be cancelled. So it is 1/r^-3=125/8
    If that's the case how did it become positive later on? (r^-3 became r^3 at some point of the equation)
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    Re: Basic Algebra Question

    Quote Originally Posted by Katsumi View Post
    I failed so many tests in high school because of that. If i'm thinking in the right way below then my problem was believing that 1/x^y automatically converted into just x^y without any further simplification needed.

    So fundamentally when you have r^-3 = -125/-8 you complete the following process

    1. As r has a negative coefficient you first convert it into a fraction so that it can become positive
    -1/r^-3 = -125/-8

    2. As both sides of the equation are negative - the negative exponents cancel out and the equation stays balanced
    1/r^3 = 125/8

    3. You then need to separate r^3 from the 1 which it is dividing. You do this by multiplying across the 8.
    8/r^3 = 125

    4. You then need to isolate variable r. As the coefficients 8 and 125 are on different sides of the equation this can be done in 2 operations.

    Operation One
    8 = 125 * r^3

    Operation Two
    r^3 = 8/125

    5. Now that variable r has been isolated we can find the value by taking the cube root of both sides of the equation.

    r = 2/5

    Is this the correct way of thinking or is there a concept behind it.

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    Re: Basic Algebra Question

    Quote Originally Posted by Katsumi View Post
    If that's the case how did it become positive later on? (r^-3 became r^3 at some point of the equation)
    The -125/(-8) is actually a positive quantity and can immediately be simplified to 125/8 (negative signs on top and bottom "cancel out"). So the original equation is equivalent to r-3 = 125/8.

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    Re: Basic Algebra Question

    Quote Originally Posted by InteGrand View Post
    ....... does that mean that a negative number divided by a negative number is a positive number

    you're shitting me if that's the case
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    Re: Basic Algebra Question

    Quote Originally Posted by Katsumi View Post
    ....... does that mean that a negative number divided by a negative number is a positive number
    Correct, a negative divided by a negative is a positive.

    (Also a negative times a negative is a positive.)

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    Re: Basic Algebra Question

    Quote Originally Posted by Katsumi View Post
    ....... does that mean that a negative number divided by a negative number is a positive number

    you're shitting me if that's the case
    That is correct, -125/-8 is the same as -1/-1 multiplied by 125/8.. where -1/-1 =1 so 125/8 multiplied by 1 is 125/8.
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    Re: Basic Algebra Question

    I never knew that..... I sat General Math for 2 years and was never told that even once.....

    I finally understand basic algebra.

    Thanks guys; very much appreciated
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    Re: Basic Algebra Question

    Quote Originally Posted by Katsumi View Post
    I never knew that..... I sat General Math for 2 years and was never told that even once.....

    I finally understand basic algebra.

    Thanks guys; very much appreciated
    Why is negative divide by negative equal to a positive?

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    Cool Re: Basic Algebra Question

    Quote Originally Posted by He-Mann View Post
    Why is negative divide by negative equal to a positive?
    Firstly, you should know that anything divided by itself is 1 so that's one way i.e.
    5/5 = 1 and -10/-10 =1

    Another way is factorisation i.e.
    check the attachement cos i cant latex
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    Re: Basic Algebra Question

    Quote Originally Posted by Mathew587 View Post
    Firstly, you should know that anything divided by itself is 1 so that's one way i.e.
    5/5 = 1 and -10/-10 =1

    Another way is factorisation i.e.
    check the attachement cos i cant latex
    Capture.PNG
    I highlighted the word 'understand' because I demanded a concrete understanding, not abstract.

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    Re: Basic Algebra Question

    CodeCogsEqn.gif

    there's three approaches in the link. i'm sure that you undestand atleast one of them
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    Re: Basic Algebra Question

    If you know a bit of abstract algebra, you can read the top answer here about why (-x)(-y) = xy in any ring (though in that case you may have already thought about this and had it answered before): http://math.stackexchange.com/questi...rs-is-positive .

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    Re: Basic Algebra Question

    Quote Originally Posted by Mathew587 View Post
    CodeCogsEqn.gif

    there's three approaches in the link. i'm sure that you undestand atleast one of them
    Still abstract. Can you provide a concrete example demonstrating this idea? i.e. incorporating realistic objects like monkey noses, hexagonal-shaped dishes, dollar coins covered in acetic acid, etc

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    Re: Basic Algebra Question

    Let x= 1 dollar coin covered in aceitic acid
    Let y= he-mann's shitty sarcasm
    -x/-x = -1(x)/-1(x) = x/x = 1
    -x/-y = -1(x)/-1(y) = x/y
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    Re: Basic Algebra Question

    I'll give you a basic example.

    Let x = Matthew587's debt to He-Mann = 2017 apologetic letters typesetted in LaTeX and must include a picture of a golden coin covered in acetic acid.
    Let y = 2017 letters sent to He-Mann.

    Then, x/y = 2017/2017 = 1.

    That is, your debt can be paid in one payment!

    Here, we are working with positives. Try incorporating negatives and try to make sense out of it.

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