# 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.

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 |2x − 1| > 5.

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

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

|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.

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

-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

-1<(2x-9)/4<1

-4<2x-9<4

5<2x<13

2.5<x<6.5

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

|(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

-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.

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