Inequalities – Inequality with absolute value – Exercise 1852

Exercise

Solve the inequality:

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

Final Answer

Show final answer

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.

We start with the first 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

$2< x\leq 5$

Hence, our final answer is

$-5\leq x\leq 5$

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