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    I'm churning through this answer paper, and I found a small step which I don't quite understand! D:

    Where in the world did the one on the top go? Why did they switch place?
    Thanks in advance! I may just be missing something obvious
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    (Original post by MarcusThePotato)
    I'm churning through this answer paper, and I found a small step which I don't quite understand! D:

    Where in the world did the one on the top go? Why did they switch place?
    Thanks in advance! I may just be missing something obvious
    If you think about the rule that any number or function over itself is 1 (\frac{f(x)}{f(x)}=1) then it stands that:

    \frac{1}{\frac{11-6x}{2x-3}}\cdot \frac{f(x)}{f(x)} = \frac{1\cdot f(x)}{\frac{11-6x}{2x-3}\cdot f(x)} = \frac{1}{\frac{11-6x}{2x-3}}

    What you can do here is set f(x)=2x-3 which means you multiply the numerator and denominator by f(x). This means you end up with 1 \cdot (2x-3) = 2x-3 on the top and \frac{11-6x}{2x-3}\cdot(2x-3)=11-6x on the bottom!
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    I always imagine it like this. You have a fraction:

    a = \frac{1}{(\frac{1}{b})}

    Multiply it by b/b (i.e. 1)

    a = \frac{b}{b} \times \frac{1}{(\frac{1}{b})}

    When multiplying fractions you multiply both the top and bottom, so it becomes:

    a = \frac{b}{b(\frac{1}{b})} so a = \frac{b}{(\frac{b}{b})}

    Now, b/b is 1, so that cancels out:

    a = \frac{b}{1}

    So, \frac{1}{(\frac{1}{b})} = \frac{b}{1}
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    bless both of you!
 
 
 
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