Square root and Approximations

Linear inequalities with absolute sign

Linear inequalities with absolute sign are not as easy as simple linear inequalities. They can be divided into 4 cases depending on the inequality sign and positive or negative right side. All these cases are described in a table given below.

Case 1

Case 2

Case 3

Case 4

$\dpi{150} \mathbf{\left | x \right |}< \mathbf{a}$ or

$\dpi{150} \mathbf{\left | x \right |}\leq \mathbf{a}$

$\dpi{150} \mathbf{\left | x \right |}> \mathbf{a}$ or

$\dpi{150} \mathbf{\left | x \right |}\geq \mathbf{a}$

$\dpi{150} \mathbf{\left | x \right |}\leq \mathbf{-a}$

$\dpi{150} \mathbf{\left | x \right |} > \mathbf{-a}$

$\dpi{120} -a < x < a$ or

$\dpi{120} -a\leq x\leq a$

After rewriting like this, we solve this compound inequality using rules of inequality to get x all by itself.

x > a or x < -a

$\dpi{120} x \geq a$ or $\dpi{120} x\leq -a$

After breaking the inequalities like this, expression can be solved further to get x by itself.

Since absolute value is always positive, it can’t be less than or equal to any negative number so this case has no solution.

Since absolute value is always positive ,it would always be greater than any negative number so this case has ‘All real numbers ’ as solution.

Lets work on some examples representing all the 4 cases to understand them better.

Example1. Solve the following inequality

$\dpi{120} {\color{Red} \left | 2x+3 \right |\leq 15}$

Solution: This is of the form $\mathbf{\left | x \right |}\leq \mathbf{a}$ (case 1) , so rewriting the inequality we get,

$\dpi{120} -15\leq 2x+3\leq 15$

-3 -3 -3 subtracting 3 from all sides, we get ,

$\dpi{120} -18 \leq 2x\leq 12$

$\dpi{120} \frac{-18}{2}\leq \frac{2x}{2}\leq \frac{12}{2}$

$\dpi{120} -9\leq x\leq 6$

In interval notation this solution can be written as [-9,6]

Example2. Solve the following inequality and write the solution in interval notation.

3|2x+1| > 9

Solution: First we isolate the absolute part. So divide both sides by 3 we get, |2x+1| > 3

This is of the form $\mathbf{\left | x \right |}> \mathbf{a}$ (case2) , so rewriting the inequality we get,

2x+1 > 3 or 2x+1 < -3

2x > 2 2x < -4

x > 1 or x < -2

Combining these two inequalities by replacing ‘or’ with U (union symbol) to write final answer in interval notation we get,

$\dpi{120} \left ( -\infty ,-2 \right )\cup \left ( 1,\infty \right )$

Example3. Solve the following inequality and write the solution in interval notation.

$\dpi{120} 8+\left | 2x+5 \right |\leq 5{\color{Red} }$

Solution: Subtract 8 from both sides to isolate absolute part we get,

$\dpi{120} \left | 2x+5 \right |\leq -3$

This is of the form $\mathbf{\left | x \right |}\leq \mathbf{-a}$ (case 3).

Since absolute value is always positive, so it can’t be less than or equal to any negative number. Therefore this inequality has No Solution.

Example4. Solve the following inequality and write the solution in interval notation.

$\dpi{120} 8+\left | 2x+5 \right |>5$

Solution: Subtract 8 from both sides to isolate absolute part we get,

$\dpi{120} \left | 2x+5 \right |> -3$

This is of the form $\mathbf{\left | x \right |} > \mathbf{-a}$ (case 4).

Since absolute value is always positive, so it is always greater than any negative number. Therefore this inequality has ‘All Real numbers ’ as solution which can be written in interval notation as, (-∞,∞)

Inequalities with absolute sign having variables on both sides:

Inequalities having more than one absolute sign or having variable on both sides, are solved using different approach. We get the intervals on a number line according to expression inside absolute signs and then solve inequalities for each interval. Lets work on some examples to understand this process better.

Example5. Solve the given inequality and write the answer in interval notation.

|x-1|+|x-2| ≥ 4

Solution: First we set the expressions inside absolute signs =0 to get points on number line.

So x-1=0 => x=1 and x-2=0 => x=2

Using these two points, we get the intervals on number line as (-∞,1) ,(1,2) and (2,∞). Here we also make use of definition of absolute function.

$\dpi{120} \left | x-1 \right |= \left\{\begin{matrix} -(x-1) & x < 1 \\ (x-1) &x\geq 1 \end{matrix}\right.$ $\dpi{120} \left | x-2 \right |= \left\{\begin{matrix} -(x-2) & x < 2 \\ (x-2) &x\geq 2 \end{matrix}\right.$

For interval (-∞,1) we have the inequality as,

-(x-1)-(x-2) ≥ 4

-x+1-x+2 ≥ 4

-2x +3 ≥ 4

-2x ≥ 1

x ≤ -½ —–(i)

For interval (1,2) we have the inequality as,

(x-1) –(x-2) ≥ 4

x-1-x+2 ≥ 4

1 ≥ 4 (which is absurd )

For interval (2,∞), we have the inequality as,

(x-1) +(x-2) ≥ 4

2x -3 ≥ 4

2x ≥ 7

$\dpi{120} x\geq \frac{7}{2}$ ——–(ii)

Combining results (i) and (ii) we get the final solution in interval form as, (-∞ , -½] U [ 7/2 ,∞)

Example6. Solve the inequality |2x-3| > x-1

Solution: Taking positive and negative cases separately,

-(2x-3) > x-1 or 2x-3 > x-1

-2x+3 > x-1 2x-x > -1+3

-2x-x > -1-3 x > 2

-3x > -4

x < 4/3

Combining these two inequalities, we get the final answer in interval form as, (-∞, 4/3) U (2,∞)

Practice problems:

Solve the following inequalities and write the solution in interval form.

1) |3x-2| ≤ ½

2) |4-x| +1 > 3

3) |x-1|+|x-2|+|x-3|≥ 6

4) 4+|x-3| < 1

Answers:

1) [1/2 , 5/6]

2) (-∞,2) U(6,∞)

3) (-∞,0] U [4,∞)

4) No solution

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