# 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) > -6Multiply by 2x - 1 throughout the equation 1 > -6(2x - 1)1 < -12x + 6Subtract 6 on both sides1 - 6 < -12x + 6 - 6-5 < -12x Divide by -12 on both sides5/12 > x 1/(2x-1) < 6Multiply by 2x - 1 throughout the equation 1 > 6(2x - 1)1 < 12x - 6Add 6 on both sides1 + 6 < 12x - 6 + 67 < 12x Divide by 12 on both sides7/12 < 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 ≤ −7Divide by -3 on both sidesx ≥ 7/3 −3(-x) + 5 ≤ −2 3x + 5 ≤ −2 Subtract 5 on both sides3x + 5 - 5 ≤ −2 - 53x ≤ −7Divide by 3 on both sidesx ≤ −7/3

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

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