Inequalities involving absolute value :
Here we shall learn to solve inequalities involving absolute value.
Given inequalities |
Solution | |
|x - a| < r |
-r < x - a < r Solve for x, by adding "a" throughout the equation -r + a < x < r + a Hence the solution is (a - r, a + r) | |
|x - a| > r |
x - a < -r (or) x - a > r Solve for x, by adding "a" throughout the equation x - a < -r (or) x - a > r x < -r + a (or) x > r + a Hence the solution is (∞, a - r) U (a + r, ∞) | |
|x - a| ≤ r |
-r ≤ x - a ≤ r Solve for x, by adding "a" throughout the equation -r + a ≤ x ≤ r + a Hence the solution is (∞, a - r] U [a + r , ∞) | |
|x - a| ≥ r |
x - a ≤ -r (or) x - a ≥ r Solve for x, by adding "a" throughout the equation x - a ≤ -r (or) x - a ≥ r x ≤ -r + a (or) x ≥ r + a Hence the solution is (∞, a - r] U [a + r, ∞) | |
|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. |
Let us look into some examples to understand the above concept.
Question 1 :
(i) |3 - x| < 7
(ii) |4x - 5| ≥ -2
(iii) |3 - (3x/4)| ≤ 1/4
(iv) |x| - 10 < -3
Solution :
(i) |3 - x| < 7
-7 < (3 - x) < 7
Subtract 3 on both sides
-7 - 3 < 3 - x - 3 < 7 - 3
-10 < -x < 4
Multiply by negative sign throughout the equation
10 > x > -4
-4 < x < 10
(ii) |4x - 5| ≥ -2
Solution :
By comparing the given question with the general form |x-a|≥r, we come to know that the value of r is negative.
Since we have modulus function on the left, we always get only positive values by applying any positive or positive value for x.
Hence the solution is all real numbers.
(iii) |3 - (3x/4)| ≤ 1/4
Solution :
-1/4 ≤ 3 - (3x/4) ≤ 1/4
(-1/4) - 3 ≤ 3 - (3x/4) - 3 ≤ (1/4) - 3
(-1 - 12)/4 ≤ - (3x/4) ≤ (1 - 12)/4
-13/4 ≤ - (3x/4) ≤ -11/4
Multiply by 4 throughout the equation, we get
-13 ≤ - 3x ≤ -11
Divide by -3 throughout the equation
13/3 ≥ x ≥ 11/3
Hence the solution is 11/3 ≤ x ≤ 13/3
(iv) |x| - 10 < -3
Solution :
x - 10 < -3 Add 10 on both sides x - 10 + 10 < -3 + 10 x < 7 |
-x - 10 < -3 Add 10 on both sides -x - 10 + 10 < -3 + 10 -x < 7 Multiply by -1 on both sides x > -7 |
Hence the solution is -7 < x < 7
Let us see the next example problem on "Inequalities Involving Absolute Value".
Question 2 :
Solve (1/|2x - 1|) < 6 and express the solution using interval notation.
Solution :
(1/|2x - 1|) < 6
Taking reciprocals on both sides, we get
|2x - 1| < (1/6)
-1/6 < 2x - 1 < 1/6
Add 1 though out the inequality, we get
(-1/6) + 1 < 2x < (1/6) + 1
(5/6) < 2x < (7/6)
Divide by 2 throughout the inequality, we get
5/12 < x < 7/12
By writing using interval notation, we get
(-∞, 5/12) U (7/12, ∞)
Question 3 :
Solve −3|x| + 5 ≤ −2 and graph the solution set in a number line
Solution :
−3|x| + 5 ≤ −2
−3(x) + 5 ≤ −2 -3x + 5 ≤ −2 -3x ≤ −2 - 5 -3x ≤ −7 x ≤ 7/3 |
−3(-x) + 5 ≥ −2 3x + 5 ≥ −2 3x ≥ −2 - 5 3x ≥ −7 x ≥ −7/3 |
(-∞, -7/3] U [7/3, ∞)
Let us see the next example problem on "Inequalities Involving Absolute Value".
Question 4 :
Solve 2|x + 1| − 6 ≤ 7 and graph the solution set in a number line.
Solution :
2|x + 1| − 6 ≤ 7
2(x + 1) − 6 ≤ 7 2x + 2 - 6 ≤ 7 2x - 4 ≤ 7 2x ≤ 7 + 4 2x ≤ 11 x ≤ 11/2 |
-2(x + 1) − 6 ≥ 7 -2x - 2 - 6 ≥ 7 -2x - 8 ≥ 7 -2x ≥ 7 + 8 -2x ≥ 15 x ≥ -15/2 |
Hence the solution is
-15/2 ≤ x ≤ 11/2
Question 5 :
Solve (1/5) |10x − 2| < 1.
Solution :
(1/5) |10x − 2| < 1
|10x - 2| < 5
-5 < 10x - 2 < 5
Add 2 through out the inequality, we get
-3 < 10x < 7
Divide by 10 through out the inequality, we get
-3/10 < x < 7/10
Question 6 :
Solve |5x − 12| < −2
Solution :
By applying the any values for x in the given inequality, we get only positive values.
So, the given condition is not possible for any vlaues of x.
Hence no solution.
After having gone through the stuff given above, we hope that the students would have understood "Inequalities Involving Absolute Value"
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