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The problem is to simplify $$\left(\dfrac{1+3z}{3z}\right)$$ I know that when I have $$\dfrac{1}{x}$$ I can bring the $x$ from the denominator to the numerator by changing it to a negative power. How would I go about doing this if the numerator has multiple terms? I have thought about adding, multiplying but none of them seem correct to me. This is what I have tried: $$1+3z+3z^{-1}$$ and $$1+(3z)(3z^{-1})$$ Both of these seem incorrect to me. Am I supposed to multiply the entire numerator with $3z^{-1}$?

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2 Answers 2

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Hint: For any $a$, $b$, $c$ (where $c\neq 0$), we have $$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}.$$


By the way, $(3z)^{-1}$ is not the same thing as $3\cdot z^{-1}$, for the same reason that $(3\cdot 2)^{-1}=\frac{1}{6}$ is not the same thing as $3\cdot \frac{1}{2}=\frac{3}{2}$.

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Your "scope" of the negative exponent is off: $\quad (3z)^{-1} = 3^{-1}z^{-1} = \large \frac{1}{3z} = \large \frac {z^{-1}}{3}$

$$\left(\dfrac{1+3z}{3z}\right) = \dfrac{(1+3z)z^{-1}}{3} = \dfrac{1}{3z} + \dfrac {3z}{3z} = \dfrac 1{3z} + 1...$$

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  • $\begingroup$ nice amy :+)... $\endgroup$
    – Mikasa
    Commented Mar 1, 2013 at 12:24

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