# How do you solve this fraction?

What is the value of: $$\frac{2}{3} - \frac{1}{16}$$

My answer: $\dfrac{1}{48}$ . I believe that it's incorrect.

Three does not divide into $16$, so I cross multiplied. What am I doing wrong?

• $\frac{2}{3}-\frac{1}{16}=\frac{32}{48}-\frac{3}{48}$ – Milly Oct 17 '14 at 3:54
• They're addition, not subtraction. It is negative one over 16. Can you break it down for me? I've tried all the methods I know of for this one. Cross Multiply. I get 1 over 48... Is this correct? – kinesis Oct 17 '14 at 3:55
• you have negative numerator in \frac{-1,16} – Milly Oct 17 '14 at 3:56
• Adding negative one over 16 is the same as subtracting (positive) one over 16. – hardmath Oct 17 '14 at 3:57
• “cross multiplication” is a technique that makes sense only when you have an equation, which you don’t. – Lubin Oct 17 '14 at 4:15

$$\frac23+\frac{-1}{16}=\left(\frac23\times 1\right)\!+\!\left(\frac{-1}{16}\times 1\right)=\left(\frac23\times\frac{16}{16}\right)\!+\!\left(\frac{-1}{16}\times\frac{3}{3}\right)=\frac{32}{48}+\frac{-3}{48}=\frac{32-3}{48}=\frac{29}{48}$$

• That was the answer Wolfram gave me. Where do you derive this x 1 past the first equals sign? How does it become a numerator of 16 on the right fraction just past the second equals sign? If only I could understand how this is done. I'm really really confused with this one. Many questions come to mind. What is the significance of 16/16 past the second equals sign (sorry my \func isn't working) – kinesis Oct 17 '14 at 4:04
• Let me explain using a different problem. Let's say I have two similar cakes. The first one is divided into two equal slices and the other one is divided into three equal slices. I get one slice from the first cake and another slice from the second cake. But I can't add the slice from the first cake with the slice in the second cake because they have different sizes. So I cut each slice of the first cake into 3 and I cut each slice of the second cake into 2. Each cake now has slices of the same size (1/6 of the cake). – Joel Reyes Noche Oct 17 '14 at 4:16
• The story in my comment above can be written in a shorter way like this: $\frac12+\frac13=(\frac12\times\frac33)+(\frac13\times\frac22)=\frac36+\frac26$. Does this make things clearer? – Joel Reyes Noche Oct 17 '14 at 4:19
• I finally figured out the "what you do to the top you must do to the bottom" rule. Which may explain why you do $\frac{16}{16}$ and $\frac{3}{3}$ after the second equal sign. I am still figuring it out. – kinesis Oct 17 '14 at 22:49
• I did this one for a trial run: $\frac{2}{7} + \frac{8}{14} = (\frac{2}{7} * \frac{14}{14}) + (\frac{8}{14} * \frac{7}{7}) = \frac{28}{98} + \frac{56}{98} = \frac{84}{98} = \frac{6}{7}$ Which was correct. It looks like I can do it. Needs more practice because it is not efficient yet. – kinesis Oct 17 '14 at 22:58

\begin{align*} \frac{2}{3}+\frac{-1}{16}&=\frac{2}{3}-\frac{1}{16} \\ &=\left(\frac{16}{16}\right)\times\frac{2}{3}-\frac{1}{16}\times\left(\frac{3}{3}\right) \\ &=\frac{32}{48}-\frac{3}{48} \\ &=\frac{29}{48} \end{align*}

Since $$3$$ and $$16=2^4$$ are coprime, their least common multiple (this is the common denominator) is their product $$3\cdot 16=48$$. Hence:

$$\frac{2}{3}-\frac{1}{16}= \frac{2\color{red}{\cdot 16}}{3\color{red}{\cdot 16}}-\frac{1\color{magenta}{\cdot 3}}{16\color{magenta}{\cdot 3}}= \frac{32}{48}-\frac{3}{48}=\frac{32-3}{48}=\frac{29}{48}.$$

Now $$29$$ and $$48$$ are again coprime, so we cannot simplify further.