6.1: Evaluating Expressions

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Page ID: 45191

Victoria Dominguez, Cristian Martinez, & Sanaa Saykali
Citrus College via ASCCC Open Educational Resources Initiative

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Definition: Absolute Value

The absolute value of a real number $a$, written $|a|$, is the distance from $a$ to $0$ on a number line.

To find $|−4|$, ask: “what is the distance from $−4$ to $0$?”. Draw a number line and see that $|−4| = 4$. Similarly, $|4| = 4$, as shown in the figure below.

$clipboard_eb8163436d100cbff9fa7ef8e4f2fd943.png$

Example 6.1.1

Evaluate the following expressions:

$|8−2|− |4−7|$
$5|−3|+|−9|^2$
$\dfrac{3}{5}|6 + (−3)^3|$
$\left|\dfrac{(−2)^2 + 12}{3} +5 \right|+|−4+2|$

Solution

To evaluate $|8 − 2| − |4 − 7|$, first simplify inside the absolute value.

$\begin{array} &&|8 − 2| − |4 − 7| &\text{Given} \\ &= |6| − |− 3| &\text{Simplify inside the absolute value} \\ &= (6) − (3) &\text{Absolute value definition} \\ &= 3 & \end{array}$

First, simplify the absolute values, then apply the required arithmetic operation.

$\begin{array} &&5| − 3| + | − 9|^2 &\text{Given} \\ &= 5(3) + (9)^2 &\text{Absolute value definition} \\ &= 15 + 81 &\text{Simplify} \\ &= 96 & \end{array}$

Use order of operations ”PEMDAS” to simplify inside the absolute value.

$\begin{array} &&\dfrac{3}{5}|6 + (−3)^3| &\text{Given} \\ &=\dfrac{3}{5}|6 + (−27)| &\text{Evaluate the exponent term} \\ &= \dfrac{3}{5} − 21 &\text{Simplify inside the absolute value} \\ &= \dfrac{3}{5} (21) &\text{Absolute value definition} \\ &= \dfrac{63}{5} & \end{array}$

To evaluate the expression in this part, first apply the order of operation ”PEMDAS” inside the absolute value to simplify.

$\begin{array} &&\left|\dfrac{(−2)^2 + 12}{3} +5 \right|+|−4+2| &\text{Given} \\ &= \left|\dfrac{(4 + 12)}{3} +5 \right|+|−2| &\text{Simplify} \\ &= \left|\dfrac{16}{3} +5 \right|+|−2| &\text{Note that \(3$ is the LCD of $\dfrac{16}{3}$ and $5$. $5$ can be written as $\dfrac{5}{1}$} \\ &= \left|\dfrac{16}{3} + \dfrac{5(3)}{1(3)} \right|+|−2| &\text{Multiply numerator and denominator of $\dfrac{5}{1}$ by LCD to add the terms inside the absolute value.} \\ &=\left|\dfrac{31}{3} \right|+|−2| & \\ &= \left(\dfrac{31}{3}\right) + (2) &\text{Absolute value definition} \\ &=\dfrac{31}{3} + 2 &\text{Similar to above, $3$ is the LCD of $\dfrac{31}{3}$ and $2$. $2$ can be written as $\dfrac{2}{1}$.} \\ &= \dfrac{31}{3} + \dfrac{2(3)}{1(3)} &\text{Multiply $\dfrac{2}{1}$ by $\dfrac{3}{3}$ to add the two terms.} \\ &= \dfrac{37}{3} & \end{array} \)

Exercise 6.1.1

Evaluate the given expressions:

$|8 − 15|$
$|− 3 −12|$
$\left|− 2 + 11 − \left( −\dfrac{6}{4} \right) \right|$
$\left|−\dfrac{1 + 5}{12} − 5\right|− 1$
$|2 (5 + 6) − 20|$
$\left|\dfrac{1}{2} (21 − 5) − |(−2)^3 \right|$
$\left|−5 |− 2(−13 + 10) \right|$
$\dfrac{3}{2} \left| 12 \left( \dfrac{−7 + 17}{(6 − 2)} \right) \right| + |− (−2)|$