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So I have a term like this:

$$(-x^3y^2+xza) : (x^2y^3)$$

For simplifying, I rewrote that as a fraction, and extracted an $x$:

$$\require{cancel}\frac{\cancel{𝚡}(-x^2y^2+za)}{\cancel{𝚡}(xy^3)}$$

But that's actually all I could come up with. The $... + za$ part got me confused.

Is there any further way to simplify this? There were some answer possibilities, but none matched what I got.

  • Answer A:

$$-\frac{x}{y} + x^{-1}zay^{-3}$$

  • Answer B:

$$-\frac{y}{x}^{-1} + \frac{xy^{-3}}{za}$$

  • Answer C:

$$\frac{-x+xza}{y}$$

  • Answer D:

$$\frac{za}{y}$$

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    $\begingroup$ Could you write the possible answers? $\endgroup$ – mfl Oct 19 '17 at 7:45
  • $\begingroup$ @mfl OK! Give me a second $\endgroup$ – Max Oct 19 '17 at 7:52
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$$\frac{-x^2y^2+za}{xy^3}=\frac{-x^2y^2}{xy^3}+\frac{za}{xy^3}=\frac{-\cancel{xy^2}(x)}{\cancel{xy^2}(y)}+za(xy^3)^{-1}=-\frac{x}{y}+x^{-1}zay^{-3}$$ Hence, correct answer is Option A.

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Two approaches:

  • You can show that $D$ isn't the answer since for $x=y=z=a=1$ the terms have different values.
  • You can show that for $az \rightarrow \infty$ and $x,y$ constant your term goes to $\infty$, but the term from answer $B$ does not.
  • You can see that for $za=1$ Answer $C$ has value $0$, $x$ doesn't matter. But in the original term $(-x^3y^2+x\cdot 1) : (x^2y^3)$ the value of $x$ does matter.

Now lets go for a more analytic approach: \begin{align*} (-x^3y^2+xza) : (x^2y^3) &= \require{cancel}\frac{\cancel{𝚡}(-x^2y^2+za)}{\cancel{𝚡}(xy^3)} \\ &= \frac{-x^2y^2}{xy^3} + \frac{za}{xy^3} \\ &= \frac{-x}{y} + zax^{-1}y^{-3} \\ &= -\left(\frac{y}{x}\right)^{-1} + x^{-1}zay^{-3} \end{align*}

So $A$ is correct.

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