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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
avatar
+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
avatar+1171 
+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

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