# How do you solve and write the following in interval notation: |1 + 5x| ≤ 11?

May 1, 2018

Solution: $- 2.4 \le x \le 2$ i.e $x \in \left[- 2.4 , 2\right]$

#### Explanation:

 | 1+5 x| <= 11 ; 1 + 5 x <= 11 or 5 x <=10 or x <= 2 OR

 | 1+5 x| <= 11 ; 1 + 5 x >= -11 or 5 x >= -12 or x >= -12/5

or $x \ge - 2.4 \therefore$ Solution: $- 2.4 \le x \le 2$ i.e $x \in \left[- 2.4 , 2\right]$ [Ans]

May 1, 2018

The solution is $x \in \left[- \frac{12}{5} , 2\right]$

#### Explanation:

The equation is

$| 1 + 5 x | \le 11$

$| 1 + 5 x | - 11 \le 0$

When $1 + 5 x = 0$

$\iff$, $5 x = - 1$

$\iff$, $x = - \frac{1}{5}$

Let's build a sign chart

$\textcolor{w h i t e}{a a a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a a a}$$- \frac{1}{5}$$\textcolor{w h i t e}{a a a a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a a a}$$1 + 5 x$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a a a}$$| 1 + 5 x |$$\textcolor{w h i t e}{a a a a a a a}$$- 1 - 5 x$$\textcolor{w h i t e}{a a a a a a}$color(white)(aaa)1+5x

$\textcolor{w h i t e}{a a a a a a}$$| 1 + 5 x | - 11$$\textcolor{w h i t e}{a a a}$$- 12 - 5 x$$\textcolor{w h i t e}{a a a}$color(white)(aaa)-10+5x

Therefore,

$- 12 - 5 x \le 0$, $\iff$, $5 x \ge - 12$, $\iff$, $x \le - \frac{12}{5}$

$x \in \left(- \infty , - \frac{1}{5}\right)$

and

$- 10 + 5 x \le 0$, $\iff$, $x \le 2$

$x \in \left(- \frac{1}{5} , + \infty\right)$

The solution is

$x \in \left[- \frac{12}{5} , 2\right]$

graph{|1+5x|-11 [-16.09, 15.95, -12.05, 3.97]}

May 1, 2018

$\text{ }$

Solution to color(red)(|1+5x|<=11 is color(blue)(-12/5<= x <= 2

Interval Notation: color(green)([-12/5, 2]

#### Explanation:

$\text{ }$
color(green)("Step 1:"

Given the inequality: color(red)(|1+5x|<=11

There are two possible cases:

Case 1: color(red)(1+5x<=11 $A n d$

Case 2: color(red)(1+5x>=-11

Case 1:

$1 + 5 x \le 11$

Subtract $\textcolor{red}{1}$ from both sides of the inequality.

$1 + 5 x - \textcolor{red}{1} \le 11 - \textcolor{red}{1}$

$\cancel{1} + 5 x - \textcolor{red}{\cancel{1}} \le 11 - \textcolor{red}{1}$

$5 x \le 10$

Divide both sides of the inequality by $\textcolor{red}{5}$

$\frac{5 x}{\textcolor{red}{5}} \le \frac{10}{\textcolor{red}{5}}$

$\frac{\cancel{5} x}{\textcolor{red}{\cancel{5}}} \le {\cancel{10}}^{\textcolor{red}{2}} / \textcolor{red}{\cancel{5}}$

$\therefore x \le 2 \text{ }$ Solution 1

Case 2:

$1 + 5 x \ge - 11$

Subtract $\textcolor{red}{1}$ from both sides of the inequality.

1+5x-color(red)(1)>=-11-color(red)(1

cancel 1+5x-color(red)(cancel 1)>=-11-color(red)(1

$5 x \ge - 12$

Divide both sides of the inequality by $\textcolor{red}{5}$

$\frac{5 x}{\textcolor{red}{5}} \ge - \frac{12}{\textcolor{red}{5}}$

$\frac{\cancel{5} x}{\textcolor{red}{\cancel{5}}} \ge - \frac{12}{\textcolor{red}{5}}$

$x \ge - \frac{12}{5} \text{ }$ Solution 2

Combine both the solutions: Solution 1 and Solution 2

color(green)( :. x <=2 and x>=-12/5" " is our final solution.

You can also rewrite the combined solution as:

color(blue)(-12/5<=x<=2

In Interval Notation: color(blue)([-12/5,2]

color(green)("Step 2:"

Graph the inequality to verify our solution: Hope it helps.