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    2. Simplify these expressions:

    b) x-2 ÷ x-3

    d) (2x2 ÷ 4y3 )-2



    On 2b I get
     x^{-1} = \frac{1}{x} the answer is x

    On 2d I don't understand any of it.. what am I missing out on? Answer is
    4y6 / x4
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    For b), you can write this as (1/x2)/(1/x3) - multiplying both the numerator and the denominator by x3 gives you x /1 = x

    For d), you flip the fraction and square it.
    [(4y3)/(2x2)]2 = (16y6)/(4x4) = (4y6)/(x4)
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    (Original post by ckfeister)
    2. Simplify these expressions:

    b) x-2 ÷ x-3

    d) (2x2 ÷ 4y3 )-2



    On 2b I get
     x^-1 = \frac{1}{x} the answer is x
    Note the rule that a^{-n} = \frac{1}{a^n}.

    So here you're doing \displaystyle \frac{x^{-2}}{x^{-3}} = \frac{\frac{1}{x^2}}{\frac{1}{x^  3}} = \frac{x^3}{x^2}

    n 2d I don't understand any of it.. what am I missing out on? Answer is
    4y6 / x4
    Again, same thing here:

    \displaystyle \left(\frac{2x^2}{4y^3}\right)^{-2} = \frac{1}{\left(\frac{2x^2}{4y^3}  \right)^{2}} = \left(\frac{4y^3}{2x^2}\right)^2


    Can you take it from here knowing that \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} and (ab)^n = a^nb^n?
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    For 2b, the powers that x is raised to are both negatives. So x^-2/x^-3 becomes x^(-2--3) and since 2 negatives make a positive, it becomes x^(-2+3) which is x^1. And x^1 is obviously just x.

    For 2d, the whole bracket is raised to the power of -2. When a number is raised to a negative power, it becomes an inverse. So you get 1/(2x^2/4y^3)^2. The whole bracket is now squared. 2x^2 squared is 4x^4, and 4y^3 squared is 16y^6 - you add the indices. So then you have 1/4x^4/16y^6. Since you're dividing by a fraction, the denominator is multiplied by the numerator. So you now have 1(16y^6)/4x^4 which is just 16y^6/4x^4. You can simplify this fraction by taking out 4, so then you're left with 4y^6/x^4
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    (Original post by Zacken)
    Note the rule that a^{-n} = \frac{1}{a^n}.

    So here you're doing \displaystyle \frac{x^{-2}}{x^{-3}} = \frac{\frac{1}{x^2}}{\frac{1}{x^  3}} = \frac{x^3}{x^2}



    Again, same thing here:

    \displaystyle \left(\frac{2x^2}{4y^3}\right)^{-2} = \frac{1}{\left(\frac{2x^2}{4y^3}  \right)^{2}} = \left(\frac{4y^3}{2x^2}\right)^2


    Can you take it from here knowing that \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} and (ab)^n = a^nb^n?
    How did you get
    \displaystyle \frac{x^{-2}}{x^{-3}} = \frac{\frac{1}{x^2}}{\frac{1}{x^  3}} =\frac{x^3}{x^2} ? - this part
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    (Original post by ckfeister)
    How did you get
    \displaystyle \frac{x^{-2}}{x^{-3}} = \frac{\frac{1}{x^2}}{\frac{1}{x^  3}} =\frac{x^3}{x^2} ? - this part
    Standard GCSE fractions, \displaystyle \dfrac{\dfrac{1}{x^2}}{\dfrac{1}  {x^3}} \times \dfrac{x^3}{x^3} = \dfrac{\dfrac{x^3}{x^2}}{\dfrac{  x^3}{x^3}} = \dfrac{\dfrac{x^3}{x^2}}{1} = \dfrac{x^3}{x^2}
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    (Original post by Zacken)
    Standard GCSE fractions, \displaystyle \dfrac{\dfrac{1}{x^2}}{\dfrac{1}  {x^3}} \times \dfrac{x^3}{x^3} = \dfrac{\dfrac{x^3}{x^2}}{\dfrac{  x^3}{x^3}} = \dfrac{\dfrac{x^3}{x^2}}{1} = \dfrac{x^3}{x^2}
    Where on Earth did that x^3 come from?
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    (Original post by ckfeister)
    Where on Earth did that x^3 come from?
    Oh c'mon, you must remember that when you do things like adding fractions, you can multiply a fraction by 1.

    So for example, to do \displaystyle \frac{1}{2} + \frac{1}{4} you'd do \displaystyle \frac{1}{2} \times \frac{2}{2} = \frac{2}{2\times 2} = \frac{2}{4}.

    This is the same thing here, we're doing \displaystyle \frac{x^{-2}}{x^{-3}} \times \frac{x^3}{x^3}.

    If you don't like thinking of it like that, then what about \frac{x^{-2}}{x^{-3}} = \frac{1}{x^2} \times \frac{1}{x^{-3}} = \frac{1}{x^2} \times (x^{-3})^{-1} = \frac{x^{3}}{x^2}.

    Any help?
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    (Original post by ckfeister)
    2. Simplify these expressions:

    b) x-2 ÷ x-3

    d) (2x2 ÷ 4y3 )-2



    On 2b I get
     x^{-1} = \frac{1}{x} the answer is x

    On 2d I don't understand any of it.. what am I missing out on? Answer is
    4y6 / x4
    Name:  20161106_162245-1[1].jpg
Views: 85
Size:  155.5 KB

    Sorry about the quality of the picture! :P Also, where I've added the indices - that was a mistake. It was meant to be multiplying
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    \frac{x^{-2}}{x^{-3}}=x^{-2--3}=x^{-2+3}=x
    \left(\frac{2x^2}{4y^3}\right)^{-2}=\frac{\left(2x^2\right)^{-2}}{\left(4y^3\right)^{-2}}=\frac{\left(4y^3\right)^{2}}  {\left(2x^2\right)^{2}}=\frac{ \left( 2y^3 \right)^{2}}{\left(x^2\right)^{2  }}=\frac{4y^6}{x^4}
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    (Original post by Zacken)
    Oh c'mon, you must remember that when you do things like adding fractions, you can multiply a fraction by 1.

    So for example, to do \displaystyle \frac{1}{2} + \frac{1}{4} you'd do \displaystyle \frac{1}{2} \times \frac{2}{2} = \frac{2}{2\times 2} = \frac{2}{4}.

    This is the same thing here, we're doing \displaystyle \frac{x^{-2}}{x^{-3}} \times \frac{x^3}{x^3}.

    If you don't like thinking of it like that, then what about \frac{x^{-2}}{x^{-3}} = \frac{1}{x^2} \times \frac{1}{x^{-3}} = \frac{1}{x^2} \times (x^{-3})^{-1} = \frac{x^{3}}{x^2}.

    Any help?
    yes, thanks
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    (Original post by ckfeister)
    yes, thanks
    Cool.
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    (Original post by JustARandomer123)
    Name:  20161106_162245-1[1].jpg
Views: 85
Size:  155.5 KB

    Sorry about the quality of the picture! :P Also, where I've added the indices - that was a mistake. It was meant to be multiplying
    perfection, simple enough for a baby to understand.. good.
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    (Original post by ckfeister)
    perfection, simple enough for a baby to understand.. good.
    😂😂 glad you found it helpful
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    (Original post by otah007)
    \left(\frac{2x^2}{4y^3}\right)^{-2}=\frac{\left(2x^2\right)^{-2}}{\left(4y^3\right)^{-2}}=\frac{\left(4y^3\right)^{2}} {\left(2x^2\right)^{2}}=\frac{ \left( 2y^3 \right)^{2}}{\left(x^2\right)^{2 }}=\frac{4y^6}{x^4}
    Fixed: \left(\frac{2x^2}{4y^3}\right)^{-2}=\frac{\left(2x^2\right)^{-2}}{\left(4y^3\right)^{-2}}=\frac{\left(4y^3\right)^{2}}  {\left(2x^2\right)^{2}}=\frac{ \left( 2y^3 \right)^{2}}{\left(x^2\right)^{2  }}=\frac{4y^6}{x^4}
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    (Original post by Zacken)
    Fixed: \left(\frac{2x^2}{4y^3}\right)^{-2}=\frac{\left(2x^2\right)^{-2}}{\left(4y^3\right)^{-2}}=\frac{\left(4y^3\right)^{2}}  {\left(2x^2\right)^{2}}=\frac{ \left( 2y^3 \right)^{2}}{\left(x^2\right)^{2  }}=\frac{4y^6}{x^4}
    Thanks, I'll add it in.
 
 
 
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