# 4. Solving Inequalities with Absolute Values

For inequalities involving absolute values ie. |*x*|, we use the following relationships, for some number *n*:

If `|f(x)| > n`, then this means:

`f(x) < -n` or `f(x) > n`

If `|f(x)| < n`, then this means:

`-n < f(x) < n`

### Example 1

Solve the inequality |*x* − 3| < 2.

Answer

Applying the relationships discussed above:

`- 2 < x - 3 < 2`

Adding `3` to all sides, we get:

`-2 + 3 < x - 3 + 3 < 2 + 3`

`1 < x < 5`

Here is the graph of our solution:

Convince yourself that this answer is correct by checking. Try `x = 0` (should fail, because it is outside the range of our answer), `x = 3` (should work) and `x = 10` (should fail). Every time you check like this, it becomes clearer why we solve it this way.

### Example 2

Solve the inequality |2*x* − 1| > 5.

Answer

Applying the relationships discussed earlier:

`2x - 1 < -5 \ or \ 2x - 1 > 5`

Solving both inequalities, we get:

`2x < -5 + 1` or `2x > 5 + 1` `2x < -4` or `2x > 6` `x < -2 ` or `x > 3 `

Here's the graph of our solution:

### Example 3

Solve the inequality `2|(2x)/3 + 1|>=4`

Answer

`2|(2x)/3 + 1|>=4`

This gives

`|(2x)/3 + 1|>=2`

So either...

`(2x)/3+1<=-2`

OR

`(2x)/3+1>=2`

Multiply both sides by 3

`2x + 3 ≤ -6 ``2x ≤ -9`

`x<=-9/2`

OR

`2x + 3 ≥ 6 ``2x ≥ 3`

`x>=3/2`

Solution: `x<=-9/2 = -4.5,\ "or"\ x>=3/2=1.5`

Here's the graph of our solution:

### Example 4

Solve the inequality `|3 − 2x| < 3`

Answer

`|3 − 2x| < 3`

`-3 < 3 - 2x < 3`

Subtract `3` from all sides:

`-6 < -2x < 0`

Divide all sides by `-2`

`3 > x > 0`

(note the change in sense due to dividing by a negative number)

A better way to write this is: `0 < x < 3`, since we usually have smaller numbers on the left hand side.

Here's the solution graph:

### Example 5

A technician measures an electric current which is `0.036\ "A"`
with a possible error of `±0.002\ "A"`. Write this
current, *i*, as an inequality with absolute
values.

Answer

The possible error of `± 0.002\ "A"` means that the **difference** between the actual current and the value of
`0.036\ "A"` **cannot be more than** `0.002\ "A"`.

So the values of *i* we have can be expressed as:

`0.034 ≤ i ≤ 0.038`

We can simply write this as:

`|i − 0.036 | ≤ 0.002`

Here's the solution graph:

### Exercises

**1. **Solve ` |5 − x| ≤ 2`

Answer

`-2 ≤ 5 - x ≤ 2`

`-7 ≤ - x ≤ -3`

`7 ≥ x ≥ 3`

It is better to write this as: `3 ≤ x ≤ 7`.

Here's the graph of the solution:

**2. **Solve* *`|(2x-9)/4|<1`

Answer

**3.** Solve `|(4x)/3-5|>=7`

Answer

`|(4x)/3-5|>=7`

`(4x)/3-5<=-7` | OR | `(4x)/3-5>=7` |

`4x-15<=-21` `4x<=-6` `x<=-1.5` |
`4x-15>=21` `4x>=36` `x>=9` |

So the solution is: `x ≤ -1.5` or `x ≥ 9`.

Here's the graph of the solution:

**4. **Solve* *`|x^2+3x-1|<3`

Answer

`-3<x^2+3x-1<3`

Now we need to break this up into 2 parts:

#### Left part

`-3<x^2+3x-1`

`0<x^2+3x+2`

`x^2+3x+2>0`

`(x+2)(x+1)>0`

Critical values for the left side are:

`x = -2 and x = -1`.

This gives:

`x < -2 and x > -1`.

#### Right part

`x^2+3x-1<3`

`x^2+3x-4<0`

`(x+4)(x-1)<0`

Critical values:

`x = -4` and `x = 1`

This gives:

`-4 < x < 1`

It's easier to graph these together on one axis to decide the final result:

These 3 regions intersect in the following 2 places:

This gives us our final solution: −4 < *x* < −2 and −1 < *x* < 1.

Remember to **check** numbers inside and outside the given solution region to make sure it works.