# How do you solve the inequality abs(3-x)<=3?

Mar 29, 2017

$0 \le x \le 6$

#### Explanation:

$\text{Given the inequality } | x | \le a$

$\text{the solution is always of the form } - a \le x \le a$

$\Rightarrow - 3 \le \textcolor{red}{3 - x} \le 3$

Isolate x in the centre interval while obtaining numeric values in the 2 end intervals.

subtract 3 from ALL intervals.

$- 3 - 3 \le \cancel{3} \cancel{- 3} - x \le 3 - 3$

$\Rightarrow - 6 \le - x \le 0$

multiply by - 1 to obtain x

$\textcolor{b l u e}{\text{Note}}$ when multiplying/dividing an inequality by a $\textcolor{b l u e}{\text{negative}}$ value we must $\textcolor{red}{\text{ reverse the inequality symbol}}$

$\Rightarrow 6 \ge x \ge 0 \leftarrow \textcolor{red}{\text{ reverse symbol}}$

$\Rightarrow x \le 6 \textcolor{red}{\text{ and }} x \ge 0$

$\Rightarrow 0 \le x \le 6 \text{ is the solution}$

$x \in \left[0 , 6\right] \leftarrow \textcolor{red}{\text{ in interval notation}}$

Mar 29, 2017

0≤x≤6, $x \setminus \in \left[0 , 6\right]$
We can rewrite the equation as -3≤3-x≤3. Subtract both sides by $3$ to get -6≤-x≤0. Now, we multiply both sides by $- 1$ and flip the inequality signs: 6≥x≥0. This can be rewritten to 0≤x≤6.
Other notations for this answer: $x \setminus \in \left[0 , 6\right]$