###### Example3.5.3

Suppose we want to know the total cost for a box of cereal that weighs $$18$$ ounces, assuming it costs the same per ounce as the $$21$$-ounce box. Letting $$C$$ be this unknown cost (in dollars), we could set up the following proportion:

\begin{align*} \frac{\text{cost in dollars}}{\text{weight in oz}}\amp=\frac{\text{cost in dollars}}{\text{weight in oz}}\\ \frac{\$3.99}{21\,\text{oz}}\amp=\frac{\$C}{18\,\text{oz}} \end{align*}

To solve this proportion, we will first note that it will be easier to solve without units:

\begin{equation*} \frac{3.99}{21}=\frac{C}{18} \end{equation*}

Next we want to recognize that each side contains a fraction. Our usual approach for solving this type of equation is to multiply each side by the least common denominator (LCD). In this case, the LCD of $$21$$ and $$18$$ is $$126\text{.}$$ As with many other proportions we solve, it is often easier to just multiply each side by the common denominator of $$18\cdot 21\text{,}$$ which we know will make each denominator cancel:

\begin{align*} \frac{3.99}{21}\amp=\frac{C}{18}\\ \multiplyleft{18\cdot21}\frac{3.99}{21}\amp=\frac{C}{18}\multiplyright{18\cdot21}\\ 18\cdot\cancelhighlight{21}\frac{3.99}{\cancelhighlight{21}}\amp=\frac{C}{\cancelhighlight{18}}\cdot\cancelhighlight{18}\cdot21\\ 71.82\amp=21C\\ \divideunder{71.82}{21}\amp=\divideunder{21C}{21}\\ C\amp=3.42 \end{align*}

So assuming the cost is proportional to the cost of the $$21$$-ounce box, the cost for an $$18$$-ounce box of cereal would be $$\3.42\text{.}$$

in-context