Recall that the absolute value of a number represents the distance of that number from the origin.

Let's take a look at the number line below.

Since -3 and 3 are at the same distance from the origin, we have

|-3| = |3| = 3.

In general, for any real number a,

|-a| = |a|.

We can use this fact to simplify expressions containing absolute value. Let's see an example.

FLAG

Simplify the expression |-2x|.

EXPLANATION

Since |-a| = |a|, we can rewrite our expression as follows:

\begin{align*} |-2x| &= |2x| \end{align*}

FLAG

$|-4x|=$

a
$-|4x|$
b
$16x$
c
$-4$
d
$|4x|$
e
$4$

$\left| -\dfrac{4z}{7} \right| =$

a
$- \dfrac{4}{7}$
b
$-\left| \dfrac{4z}{7} \right|$
c
$\left| \dfrac{4z}{7} \right|$
d
$\dfrac{4}{7}$
e
$\left| \dfrac{4}{7} \right|$

The absolute value operation is distributive with respect to multiplication. This means that the absolute value of a product equals the product of the absolute values.

To demonstrate, consider the following expression:

|(-6) \cdot 2|

We can evaluate this expression in two separate ways:

  • The first is to find the absolute value of the product. In doing this, we get

\begin{align*} |(-6) \cdot 2| &= |-12| \\[5pt] &= 12. \end{align*}

  • The second is to find the product of the absolute values. In doing this, we get

\begin{align*} |(-6) \cdot 2| &=|-6| \cdot |2| \\[5pt] &= 6 \cdot 2\\[5pt] &= 12. \end{align*}

In general

|a \cdot b| = |a|\cdot |b|.

The absolute value operation is also distributive with respect to division. In general,

\left|\dfrac{a}{b}\right| = \dfrac{|a|}{|b|}.

FLAG

Simplify the expression |-18a|.

EXPLANATION

Firstly, since |-a| = |a|, we can write our expression as

|18a|

Then, we distribute the absolute value over the multiplication, and simplify:

\begin{align*} |18a| &= \\[5pt] |18 \cdot a| &= \\[5pt] |18| \cdot |a| &= \\[5pt] 18 \cdot |a| &= \\[5pt] 18|a| \end{align*}

FLAG

$|5x|=$

a
$5$
b
$-5|x|$
c
$-5$
d
$5|x|$
e
$-|5x|$

Simplify the expression $|-11x|.$

a
$-11$
b
$11|x|$
c
$-|11x|$
d
$-11|x|$
e
$11$

Simplify the expression \left|-\dfrac{9}{5c}\right|.

EXPLANATION

Firstly, since |-a| = |a|, we can write our expression as

\left|\dfrac{9}{5c}\right|.

Then, we distribute the absolute value over the division, and simplify:

\begin{align*} \left|\dfrac{9}{5c}\right| &=\\[5pt] \dfrac{|9|}{|5c|} &= \\[5pt] \dfrac{9}{|5c|} &= \\[5pt] \dfrac{9}{|5| \cdot |c|} &= \\[5pt] \dfrac{9}{5|c|} \end{align*}

FLAG

$\left|-\dfrac{3}{2x}\right|=$

a
$6x$
b
$-\dfrac{3}{2|x|}$
c
$\dfrac{3}{2|x|}$
d
$\dfrac{3}{2}$
e
$-6x$

Simplify the expression $\left|-\dfrac{5x}{3y}\right|.$

a
$\dfrac{5}{3}$
b
$15xy$
c
$\dfrac{5|x|}{3|y|}$
d
$-15xy$
e
$-\dfrac{5|x|}{3|y|}$
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