How do you solve this fraction?

Question

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?

Answer 1

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

Answer 2

\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*}

Answer 3

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.