# How do you solve and graph abs(2c-1)<=7?

Jan 11, 2018

$- 3 \setminus \le c \setminus \le 4$

#### Explanation:

$| 2 c - 1 | \setminus \le 7$

Consider a general inequality with the absolute value:
$| f \left(x\right) | \setminus \le a$

By definition this is equivalent to solving the following inequalities:
$f \left(x\right) \setminus \le a$ and $f \left(x\right) \setminus \ge - a$

In our specific case this means:
$2 c - 1 \setminus \le 7$
$2 c - 1 \setminus \ge - 7$

So let's solve the first one:
$2 c - 1 \setminus \le 7$
$\setminus \textcolor{b l u e}{\left(1\right)} + 2 c - 1 \setminus \le 7 + \setminus \textcolor{b l u e}{\left(1\right)}$
$2 c \setminus \le 8$
$\setminus \frac{2}{2} c \setminus \le \setminus \frac{8}{2}$
$c \setminus \le 4$

Now the second:
$2 c - 1 \setminus \ge - 7$
$2 c \setminus \ge - 6$
$c \setminus \ge - 3$

This is the graph of the first inequality:
graph{x <= 4 [-6.244, 6.243, -3.12, 3.123]}

And this of the second:
graph{x => - 3 [-4.163, 1.997, -1.455, 1.624]}

So the solution and the final graph are the common part of the two inequalities:
$- 3 \setminus \le c \setminus \le 4$