###### Example1.4.9

Use the order of operations to simplify the following expressions.

1. $$4-3\abs{5-7}\text{.}$$ For this expression, we'll treat the absolute value bars just like we treat parentheses. This implies we'll simplify what's inside the bars first, and then compute the absolute value. After that, we'll multiply and then finally subtract:

\begin{align*} 4-3\abs{5-7} \amp= 4-3\abs{\overbrace{5-7}}\\ \amp= 4-3\overbrace{\abs{\highlight{-2}}}\\ \amp= 4-\overbrace{3(\highlight{2})} \\ \amp= 4-\highlight{6}\\ \amp= \highlight{-2} \end{align*}

We may not do $$4-3=1$$ first, because $$3$$ is connected to the absolute value bars by multiplication (although implicitly), which has a higher order than subtraction.

2. $$8-\sqrt{5^2-8\cdot 2}\text{.}$$ This expression has an expression inside the radical of $$5^2-8\cdot 2\text{.}$$ We'll treat this radical like we would a set of parentheses, and simplify that internal expression first. We'll then apply the square root, and then our last step will be to subtract that expression from $$8\text{:}$$

\begin{align*} 8-\sqrt{5^2-8\cdot 2} \amp= 8-\sqrt{\overbrace{5^2}-8\cdot 2}\\ \amp= 8-\sqrt{\highlight{25}-\overbrace{8\cdot 2}}\\ \amp= 8-\sqrt{\overbrace{25-\highlight{16}}}\\ \amp= 8-\overbrace{\sqrt{\highlight{9}}}\\ \amp= 8-\highlight{3}\\ \amp= \highlight{5} \end{align*}
3. $$\dfrac{2^4+3\cdot 6}{5-18\div 2}\text{.}$$ For this expression, the first thing we want to do is to recognize that the main fraction bar serves as a separator that groups the numerator and groups the denominator. Another way this expression could be written is $$(2^4+3\cdot6)\div(15-18\div 2)\text{.}$$ This implies we'll simplify the numerator and denominator separately according to the order of operations (since there are implicit parentheses around each of these). As a final step we'll simplify the resulting fraction (which is division).

\begin{align*} \frac{2^4+3\cdot 6}{5-18\div 2} \amp= \frac{\overbrace{2^4}+3\cdot 6}{5-\underbrace{18\div 2}}\\ \amp=\frac{\highlight{16}+\overbrace{3\cdot 6}}{5-\highlight{9}}\\ \amp=\frac{16+\highlight{18}}{\highlight{-4}}\\ \amp=\frac{\highlight{34}}{-4}\\ \amp=-\frac{17}{2} \end{align*}
in-context
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