# Inequalities – Inequality with absolute value – Exercise 1852

Exercise

Solve the inequality:

$$|x+2|+|x-2|\leq 10$$

$$-5\leq x\leq 5$$

Solution

$$|x+2|+|x-2|\leq 10$$

We check when the phrases in the absolute values equal zero:

$$x+2=0 \rightarrow x=-2$$

$$x-2=0 \rightarrow x=2$$

We divide the x-axis into foreign sections by the points we found. We get three sections:

$$x\leq -2, -2< x\leq 2, x>2$$

In each section, we take the following steps:

1. Choose any number in the section.
2. Get rid of the absolute values ​​by the sign according to the number we chose, and solve the inequality.
3. Intersect result with the original section.

$$x\leq -2$$

We choose the number x = -4. We set the number in our inequality and remove the absolute values by definition – If the result is positive, we simply remove the absolute value, and if the result is negative, we remove the absolute value and multiply by minus one. Hence, we get

$$-(x+2)-(x-2)\leq 10$$

We solve the inequality:

$$-x-2-x+2\leq 10$$

$$-2x\leq 10$$

$$x\geq -5$$

Now, intersect this result with the original section. Meaning,

$$x\geq -5$$

and

$$x\leq -2$$

Together we get

$$-5 \leq x\leq -2$$

Moving on to the second section:

$$-2< x\leq 2$$

We choose the number x = 0. We set the number in our inequality and remove the absolute values by definition – If the result is positive, we simply remove the absolute value, and if the result is negative, we remove the absolute value and multiply by minus one. Hence, we get

$$(x+2)-(x-2)\leq 10$$

Solve the inequality:

$$x+2-x+2\leq 10$$

$$4\leq 10$$

Hence, the inequality solution is all x.

Now, intersect this result with the original section:

$$-2< x\leq 2$$

together, we get

$$-2< x\leq 2$$

Lastly, the third section:

$$x>2$$

We choose the number x = 4. We set the number in our inequality and remove the absolute values by definition – If the result is positive, we simply remove the absolute value, and if the result is negative, we remove the absolute value and multiply by minus one. Hence, we get

$$(x+2)+(x-2)\leq 10$$

Solve the inequality:

$$x+2+x-2\leq 10$$

$$2x\leq 10$$

$$x\leq 5$$

Now, intersect this result with the original section, meaning

$$x\leq 5$$

and

$$x>2$$

together we get

$$2< x\leq 5$$

The final step is to take all the solutions we received and union them, meaning

$$-5 \leq x\leq -2$$

or

$$-2< x\leq 2$$

or

$$2< x\leq 5$$

$$-5\leq x\leq 5$$