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Find x such that \($\lfloor x \rfloor + x = \dfrac{13}{3}$\). Express x as a common fraction.

 

Thanks in advance!!!

 Jul 17, 2020
 #1
avatar+110727 
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Next time please present your question properly by deleting the dollar signs

 

\(\lfloor x \rfloor + x = \dfrac{13}{3}\\ \lfloor x \rfloor + x =4 \dfrac{1}{3}\\ \text{obviously}\;\;\lfloor x \rfloor\;\;\text{is a whole number }\\ so\\ \lfloor x \rfloor + \lfloor x \rfloor + \dfrac{1}{3} = 4 \dfrac{1}{3}\\ so\;\;\;\lfloor x\rfloor=2\\ x=2\frac{1}{3} \)

 Jul 17, 2020
edited by Melody  Jul 17, 2020
 #2
avatar+254 
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Oopsy. I think that was a typo. Thanks Melody :)

 Jul 17, 2020

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