Let's take a number, say, -3. We can depict it on a number line as follows:

The distance between the number -3 and 0 is known as the absolute value of -3. We write it as |-3| = {\color{red}3}.

Similarly, |\,2\,| , the absolute value of the number 2 , is the distance between 2 and 0 :

We write |\,2\,| = {\color{red}2}.

So, the absolute value simply

  • removes the minus sign, if the number is negative, or

  • does nothing, if the number is non-negative.

Sometimes, we can be asked to evaluate an expression involving the absolute value function, like |x +1| + 5 when x=-3. In this case, we evaluate the expression inside the absolute value before simplifying the rest of the expression: \begin{align} |x + 1| + 5 & =\\[5pt] |(-3) + 1| + 5 & =\\[5pt] |-2| + 5 & =\\[5pt] 2 + 5 & =\\[5pt] 7 & \end{align}

FLAG

Evaluate |x - 6| + 2 x +3 for x = -3.

EXPLANATION

We substitute x = -3 and then evaluate the absolute value before we simplify the rest of the expression: \begin{align} |x - 6| + 2 x +3 & =\\[5pt] |-3 - 6| + 2 (-3) + 3 & =\\[5pt] |-9| + 2(-3) + 3 & =\\[5pt] 9 + 2(-3) + 3 & =\\[5pt] 9 - 6 + 3 & =\\[5pt] 6 \end{align}

FLAG

Evaluate $6-|x-9|$ for $x=-2.$

a
$-16$
b
$5$
c
$16$
d
$-5$
e
$-2$

If $t=-1,$ then $|t^2+t-1|+t=$

a
b
c
d
e

If $p = 3,$ then $|3 - 5^{p-2}| + 7 = $

a
b
c
d
e

Evaluate |2 - x| + 1 + 5|x| for x = 5.

EXPLANATION

We substitute x = 5 and then evaluate the absolute values before we simplify the rest of the expression: \begin{align} |2 - x| + 1 + 5|x| & =\\[5pt] |2 - 5| + 1 + 5|5| & =\\[5pt] |-3| + 1 + 5(5) & =\\[5pt] 3 + 1 + 25 & =\\[5pt] 4 + 25 & =\\[5pt] 29 \end{align}

FLAG

If $t=0,$ then $|t-2| - |t+3| = $

a
b
c
d
e

If $p = 1,$ then $|p|^4 - 4|p-3|^2 + 6 =$

a
b
c
d
e

If $y=0,$ then $2|1-3^y| - |2^y - 4| = $

a
b
c
d
e
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