Here we are going to see, "How to solve inequalities with modulus".

If the given question is in any of the following forms, we have to follow the given methods to solve for "x".

Questions are in the form |
First step to be done |
Solution |

|x - a| < r |
-r < x - a < r |
(-r + a, r + a) |

|x - a| ≤ r |
-r ≤ x - a ≤ r |
[a- r, a + r] |

|x - a| > r |
x - a < -r and x - a > r |
(∞,a-r)U(a+r,∞) |

|x - a| ≥ r |
x - a ≤ -r and x - a ≥ r |
(∞,a-r]U[a+r,∞) |

Let us look into some example problems to understand the above concepts.

**Example 1 :**

Solve the absolute value inequality given below

|x - 9| < 2

and express the solution in interval notation.

**Solution :**

-2 < x - 9 < 2

Add 9 throughout the equation

-2 + 9 < x - 9 + 9 < 2 + 9

7 < x < 11

Hence the solution set of the above absolute inequality is (7, 11).

**Example 2 :**

Solve the absolute value inequality given below

|2/ (x - 4)| > 1 , x ≠ 4

and express the solution in interval notation.

**Solution :**

From the given inequality, we have that 2 > (x - 4)

-2 < x - 4 < 2

Add 4 throughout the inequality

-2 + 4 < x - 4 + 4 < 2 + 4

2 < x < 6

We cannot express the solution as (2, 6). Because in the middle of 2 and 6, we have the value 4.

So, we have to split it into two intervals.

(2, 4) U (4, 6)

**Example 3 :**

Solve the absolute value inequality given below

|3 - (3x/4)| ≤ 1/4

and express the solution in interval notation.

**Solution :**

(-1/4) ≤ 3 - (3x/4) ≤ (1/4)

(-1/4) ≤ (12 - 3x)/4 ≤ (1/4)

Multiply by 4 throughout the equation

-1 ≤ (12 - 3x) ≤ 1

Subtract 12 throughout the equation

-1 - 12 ≤ 12 - 3x - 12 ≤ 1 -12

-13 ≤ - 3x ≤ -11

Divided by (-3) throughout the equation

-13/(-3) ≤ - 3x ≤ -11

13/3 ≤ x ≤ 11/3

11/3 ≤ x ≤ 13/3

Hence the solution set of the above absolute inequality is [11/3, 13/3].

**Example 4 :**

Solve the absolute value inequality given below

|6x + 10| ≥ 3

and express the solution in interval notation.

**Solution :**

6x + 10 ≤ -3 and 6x + 10 ≥ 3

6x + 10 ≤ -3 Subtract 10 on both sides 6x + 10 - 10 ≤ -3 - 10 6x ≤ -13 Divide by 6 on both sides x ≤ -13/6 |
6x + 10 ≥ 3 Subtract 10 on both sides 6x + 10 - 10 ≥ 3 - 10 6x ≥ -7 Divide by 6 on both sides x ≥ -7/6 |

Hence the solution set of the above absolute inequality is (-∞, -13/6] U [-7/6, ∞).

After having gone through the stuff given above, we hope that the students would have understood "How to solve inequalities with modulus".

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