6.1: Evaluating Expressions
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.
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} \)
Evaluate the given expressions:
- \(|8 − 15|\)
- \(|− 3 −12|\)
- \(\left|− 2 + 11 − \left( −\dfrac{6}{4} \right) \right|\)
- \(\left|−\dfrac{1 + 5}{12} − 5\right|− 1\)
- \(|2 (5 + 6) − 20|\)
- \(\left|\dfrac{1}{2} (21 − 5) − |(−2)^3 \right|\)
- \(\left|−5 |− 2(−13 + 10) \right|\)
- \(\dfrac{3}{2} \left| 12 \left( \dfrac{−7 + 17}{(6 − 2)} \right) \right| + |− (−2)|\)