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# simplifying this operation w/ negative exponents?

0
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$${3^{-1} + 4^{-1} \over 6^{-1}}$$

so i know i can get rid of the negatives by:

$${6^{1} \over 3^{1} + 4^{1}}$$

then i got:

$${6 \over 7}$$

but the right answer is $${7 \over 2}$$, what did i do wrong?

Guest Jul 4, 2018
#1
+1

3^-1 + 4^-1 means =1/3 + 1/4. But 6^-1 also means 1/6, so:

[1/3+1/4] / 1/6=7/12 x 6/1=42/12 =7/2

Guest Jul 4, 2018
#2
+1198
+1

Do this:

$$\dfrac{3^{-1}+4^{-1}}{6^{-1}} \hspace{.1cm} =\dfrac {\dfrac {1}{3}+\dfrac {1}{4}}{6^{-1}}\\ \\\:\\ \hspace{.1cm} =\dfrac {\dfrac {7}{12}}{6^{-1}} \hspace{.1cm} =\dfrac {7}{6^{-1}\cdot 12}\\ \\\:\\ \hspace{.1cm}=\dfrac {7}{2}\\$$

GA

GingerAle  Jul 4, 2018