# How do you solve and write the following in interval notation: | 2x - 3 |<4?

Jun 24, 2016

The absolute value of a number depends on whether the number is positive, zero or negative. Then you have to take the given inequality into account.

#### Explanation:

$| a | = a$, if $a > 0$
$| a | = - a$, if $a < 0$
$| 0 | = 0$

There are then 3 cases to have a look at:
$2 x - 3 > 0$, thus $2 x > 3$, then $x > \frac{3}{2}$

In this case $| 2 x - 3 | = 2 x - 3$
and the inequality is, $2 x - 3 < 4$, so
$2 x < 4 + 3$, or $2 x < 7$, that is
$x < \frac{7}{2}$.
Thus, at the same time , $\frac{3}{2} < x$ and $x < \frac{7}{2}$, thus the elements $x$ between $\frac{3}{2}$ and $\frac{7}{2}$ satisfy the inequality.

Similarly, if $2 x - 3 < 0$, thus $2 x < 3$, then $x < \frac{3}{2}$
In this case $| 2 x - 3 | = - 2 x + 3$
and the inequality is, $- 2 x + 3 < 4$, so
$- 2 x < 4 - 3$, or $- 2 x < 1$, that is
$- 1 < 2 x$ and thus $- \frac{1}{2} < x$
Then, at the same time , $- \frac{1}{2} < x$ and $x < \frac{3}{2}$, thus the elements $x$ between $- \frac{1}{2}$ and $\frac{3}{2}$ satisfy the inequality.

Finally, when $2 x - 3 = 0$, $x = \frac{3}{2}$, and plugging the value $\frac{3}{2}$ into the inequality we get $0 < 4$, which is true, and then $x = \frac{3}{2}$ also satisfies the inequality.

Summarising, ALL the elements $x$ between $- \frac{1}{2}$ and $\frac{7}{2}$ satisfy the inequality; that is the open interval ($- \frac{1}{2}$, $\frac{7}{2}$)