SOLVING ABSOLUTE VALUE INEQUALITIES IN INTERVAL NOTATION

If the given inequalities are in the following form, we may represent the the expression inside the absolute value sign between the range -r and r and solve for x.

Given function

Solution

(i) |x - a| < r

(ii) |x - a| > r

(iii) |x - a| ≤ r

(iv) |x - a| ≥ r

(a - r, a + r)

(∞, a - r) U (a + r, ∞)

(∞, a - r] U [a + r , ∞)

(∞, a - r] U [a + r, ∞)

Here, we may have some different cases.

Given function

Solution

|x - a| > -r

(or)

|x - a| ≥ - r

Hence the solution is all real numbers. Since we have modulus sign in the left side, always we get positive value as answer.

|x - a| < -r

(or)

|x - a| ≤ - r

There is no solution. By applying any positive and negative values for x, we get only positive answer.

Because we have modulus sign in the left side.

Example 1 :

Solve 1/|2x - 1| < 6 and express the solution using the interval notation.

Solution :

Now we have to split the given inequality into two branches.

1/(2x-1) > -6

Multiply by 2x - 1 throughout the equation

1 > -6(2x - 1)

1 < -12x + 6

Subtract 6 on both sides

1 - 6 < -12x + 6 - 6

-5 < -12x

Divide by -12 on both sides

5/12 > x

1/(2x-1) < 6

Multiply by 2x - 1 throughout the equation

1 > 6(2x - 1)

1 < 12x - 6

Add 6 on both sides

1 + 6 < 12x - 6 + 6

7 < 12x

Divide by 12 on both sides

7/12 < x

In the last step of first part, we divide -12 throughout the equation. So < sign becomes > sign

Hence the solution in interval notation is (-∞, 5/12) U (7/12, ∞).

Example 2 :

Solve −3|x| + 5 ≤ −2 and graph the solution set in a number line.

Solution :

Let us split the given inequality into two parts

−3(x) + 5 ≤ −2

-3x + 5 ≤ −2

Subtract 5 on both sides

-3x + 5 - 5 ≤ −2 - 5

-3x ≤ −7

Divide by -3 on both sides

x ≥ 7/3

−3(-x) + 5 ≤ −2

3x + 5 ≤ −2

Subtract 5 on both sides

3x + 5 - 5 ≤ −2 - 5

3x ≤ −7

Divide by 3 on both sides

x ≤ −7/3

The interval notation of the above solution is (-∞, -7/3] U [7/3, ∞).

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SOLVING ABSOLUTE VALUE INEQUALITIES IN INTERVAL NOTATION

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