We are asked to find and graph the solution set for all possible values of

a

in the given inequality.

∣ a - 5 ∣ < 3

To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number less than

3

away from the midpoint in the positive direction and any number less than

3

away from the midpoint in the negative direction.

Absolute Value Inequality : Compound Inequality : ∣ a - 5 ∣ < 3 - 3 < a - 5 < 3

We can split this compound inequality into two cases, one where

a - 5

is greater than

- 3

and one where

a - 5

is less than

3 .

a - 5 > - 3 and a - 5 < 3

Let's isolate

a

in both of these cases before graphing the solution set.

a - 5 < 3

$LHS + 5 < RHS + 5$

a < 8

This inequality tells us that all values less than

8

will satisfy the inequality.

- 3 < a - 5

$LHS + 5 < RHS + 5$

2 < a

This inequality tells us that all values greater than

2

will satisfy the inequality.

The solution to this type of compound inequality is the overlap of the solution sets. Let's recombine our cases back into one compound inequality.

First Solution Set : Second Solution Set : Intersecting Solution Set : a < 8 2 < a 2 < a < 8

The graph of this inequality includes all values from $2$ to $8,$ not inclusive. We show this by using open circles on the endpoints.

Exercise