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Peace to all. While in class I was taught to solve complex fractions by an "alternative method" in which you would:

  1. Multiply the numerator and denominator by the LCD
  2. Apply the distributive property
  3. Factor and Simplify

An example: $\dfrac{4 - \dfrac 6 {x}} {\dfrac 2 {x} - \dfrac 3 {x^2}}$

  1. Multiply the numerator and denominator by the LCD (x^2):

$\dfrac{4 - \dfrac 6 {x}} {\dfrac 2 {x} - \dfrac 3 {x^2}}$

  1. Apply the distributive property:

$$\dfrac{x^2 × (4) - x^2 × \dfrac 6 {x}} {{x^2} × \dfrac 2x - x^2 × \dfrac 3 {x^2}}$$

  1. Factor and Simplify

$$\dfrac{4x^2 - 6x} {2x-3}$$

$$\dfrac{2x(2x - 3)}{2x-3}$$

A:2x

This example is very straightforward and simple. However, when I put it to use it becomes very difficult. For example for the equation: $\dfrac{\dfrac{6y}{(y + 6)}}{\dfrac{5}{(7y + 42)}}$. I get $y + 42$ as the LCD. I'm not too confident of this because when I begin to distribute I get very large numbers. How would one apply this to the prior equation?

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    $\begingroup$ Just confirm if the edit I have made in your question is correct $\endgroup$
    – user876009
    Sep 6, 2021 at 2:32
  • $\begingroup$ @JitendraSingh Peace. I don't see that option but the edit is 100% accurate. Thanks. $\endgroup$
    – יהודה
    Sep 6, 2021 at 2:34

3 Answers 3

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Hint: $7y+42=7(y+6)$. Given this, both the numerator and denominator contain $y+6$.

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$\dfrac{\dfrac{6y}{(y + 6)}}{\dfrac{5}{(7y + 42)}}$

Multiply numerator and denominator by $7y+42$

$ \dfrac{\dfrac{6y \times (7(y+6))}{y+6}}{\dfrac{5 \times (7(y+6))}{7(y+6)}} \implies \dfrac{42y}{5}$

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$\require{cancel}$ I don't know why the LCD method is used at all because it is clear, in this case, that.

$$\dfrac{\dfrac{6y}{y + 6}}{\;\dfrac{5}{7y + 42}} =\frac{6y}{\cancel{y+6}}\cdot\frac{7(\cancel{y+6})}{5} =\frac{42y}{5}$$ The LCD is needed the "example" as a means of separately making the numerator factorable and the denominator "invertible".

$$ \dfrac{4 - \dfrac 6 {x}} {\dfrac 2 {x} - \dfrac 3 {x^2}} =\dfrac{\dfrac{4x-6}{x}} {\dfrac{2x-3}{x^2}} =\dfrac{2(\cancel{2x-3})}{x} \cdot \dfrac{x^2}{\cancel{2x-3}} =2x $$

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  • $\begingroup$ Perhaps they just like to make our lives unnecessarily difficult lol. Thank you $\endgroup$
    – יהודה
    Sep 6, 2021 at 21:44
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    $\begingroup$ @יהודה Glad I could help. I see a lot of "answers" to questions that almost seem to avoid simple solutions that are obvious once seen. I hope my treatment of the "example" showed you why LCD should be invoked at all. $\endgroup$
    – poetasis
    Sep 6, 2021 at 22:04
  • $\begingroup$ No, it did help. My professor actually stated that she disapproves of this method because of how unnecessarily uncomplicated it is. Thanks again. $\endgroup$
    – יהודה
    Sep 6, 2021 at 22:06

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