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Simplify $\dfrac{18 - \dfrac 7 {3x}} {\dfrac 7 {18x} - 3}$?

I'm having a hard time simplifying this particular expression and am seeking any type of assistance in solving it.

In the expression $$\frac{18 - \dfrac 7 {3x}} {\dfrac 7 {18x} - 3}$$ I split the problems into two separate entities.

For the numerator, I get $3x$ for the LCD and then rewrite the fraction as $$54x-7\frac 7 {3x}$$ As for the denominator, I get $18x$ for the LCD and then rewrite the fraction as $$7-\frac{54x}{18x}$$

When I begin to divide, I switch the sign from division to multiplication and swap the numerator with the denominator ($7-\frac{54x}{18x}$ becomes $\frac{18x}{7-54x}$).

The product I get is $$\frac{972x^2-126x}{21x - 162x^2}$$ When I simplify I get $6-6$ which is zero. Is this answer correct?

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Using your method, the numerator $N$ is $$N=18-\dfrac{7}{3x}=\dfrac{54x}{3x}-\dfrac{7}{3x}=\dfrac{54x-7}{3x}$$ The denominator is $$D=\dfrac{7}{18x}-3=\dfrac{7}{18x}-\dfrac{54x}{18x}=\dfrac{7-54x}{18x}$$ Thus, the given fraction is $$F=N÷D=\dfrac{54x-7}{3x}÷\dfrac{7-54x}{18x}$$ $$\implies F=\dfrac{54x-7}{3x}×\dfrac{18x}{7-54x}=\boxed{-6}$$

Hope this helps. Ask anything if not clear :)

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It doesn't need to be this complicated. multilply the numerator and denominator by $18x$ leaving $$\frac{18 - \frac 7 {3x}} {\frac 7 {18x} - 3} =\frac{324x- 42} {7 - 54x}=-6$$

obviously you are a middle school student so well done for asking for help

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    $\begingroup$ First, thank you. I recently just learned this method that you utilized but even the professor stated that they don't like it so we strayed away from it. Much appreciated beloved. $\endgroup$
    – יהודה
    Sep 3, 2021 at 4:14
  • $\begingroup$ @יהודה What method does your prof prefer? How I see it, this is the most intuitive way $\endgroup$
    – hwood87
    Sep 3, 2021 at 4:20
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$\require{cancel}$ As we discussed about technique in another post

$$\dfrac{18 - \dfrac 7 {3x}} {\dfrac 7 {18x} - 3} =\dfrac{\cancel{54x-7}}{3x } \cdot \dfrac{18x}{(-1)(\cancel{54x-7})} =-6$$

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  • $\begingroup$ I have no idea when I see it done it's crystal clear but when I am trying to do it alone, its very challenging. I'm like "these exercises aren't like the examples given in the book!!!" lol. Once again thank you. $\endgroup$
    – יהודה
    Sep 6, 2021 at 23:21
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    $\begingroup$ @יהודה Do the LCD thing on the numerator and denominator separately. Then you can multiply the numerator by the inverted denominator. $\endgroup$
    – poetasis
    Sep 6, 2021 at 23:29

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