Solving Inequalities

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In a nutshell, solving inequalities is about one thing: sign changes. Find all the points at which there are sign changes – we call these points critical values.  Then determine which, if any, of the intervals bounded by these critical values result in a solution. The solution to the inequality will consist of the set of all points contained by the solution intervals. 

Method To Solve Linear, Polynomial, or Absolute Value Inequalities:

  1. Move all terms to one side of the inequality sign by applying the Addition, Subtraction, Multiplication, and Division Properties of Inequalities. You should have only zero on one side of the inequality sign.
  2. Solve the associated equation using an appropriate method. This solution or solutions will make up the set of critical values.  At these values, sign changes occur in the inequality.
  3. Plot the critical values on a number line. Use closed circles ∙ for ≤ and ≥ inequalities, and use open circles ο for < and > inequalities.
  4. Test each interval defined by the critical values. If an interval satisfies the inequality, then it is part of the solution.  If it does not satisfy the inequality, then it is not part of the solution.

Example 1 (Linear Inequality):  Solve 3x + 5(x + 1)  ≤  4x – 1 and graph the solution

3x + 5(x + 1)  ≤  4x – 1 Given

3x + 5x + 5  ≤  4x – 1 Distributive Property

8x + 5  ≤  4x – 1 Combine Like Terms

4x + 6  ≤  0 Subtract 4x from both sides, add 1 to both sides using Addition and Subtraction Properties of Inequality

Now, solve 4x + 6 = 0

4x = -6 Addition Property of Equality

x = – 6/4 = -3/2 Division Property of Equality

Plot the critical value

Test arbitrary values of each interval. You may test the value in the original inequality or it’s simplified form.  

In Interval 1, let x= 1 in 4x + 6  ≤  0.

4(1) + 6 ≤ 0 is FALSE.

In Interval 2, let x = -2 in 4x + 6 ≤  0.

4(-2) + 6 ≤ 0 is TRUE.  So Interval 2 is the solution.

The solution is x ≤ – 3/2, or in interval notation, (- ∞, – 3/2].  The graph is shown below:

Note: It is more convenient to use the circle notation for endpoints of the graph rather than the interval bracket notation since we do not know which way the brackets will point until the inequality intervals are tested.

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Example 2 (Polynomial Inequality):  Solve 3x3 + 5x2  >  4x2 and graph the solution

3x3 + 5x2  >  4x2 Given

3x3 + x2  >  0 Subtraction Property of Inequalities

Now, solve 3x3 + x2  =  0

x2(3x + 1) = 0 Distributive Property

x2 = 0 or 3x + 1 = 0 Zero Product Law

Solve x2 = 0

x=±√0 = 0 Extract Square Roots  

Solve 3x + 1 = 0

3x = -1 Add –1 to  using Addition Property of Equality

x = -1/3 Division Property of Equality

Plot the critical values, x = 0 and x =-1/3. Use open circles this time!

Test arbitrary values of each interval. You may test the value in the original inequality or it’s simplified form.  

In Interval 1, let x= 1 in x2(3x + 1) > 0.

12(3∙1 + 1) > 0 is TRUE, so Interval 1 is part of the solution.

In Interval 2, let x=-1/4 in x2(3x + 1) > 0.

(-1/4)2(3∙(-1/4) + 1) > 0 is TRUE,  since simplified, we get (1/16)(1/4) > 0,

so Interval 2 is also part of the solution.

In Interval 3, let x=-1 in x2(3x + 1) > 0.

(-1)2(3∙(-1) + 1) > 0 is FALSE, since simplified, we get (1)(-2) > 0,

so Interval 3 is NOT part of the solution.

We shade Interval 1 and Interval 2, but do not include the endpoints.

The solution is x>0 or –1/3 < x < 0.  

The interval notation of this solution is (0 , ) U (-1/3, 0).

A COMMON MISTAKE TO AVOID! 

Students often get to the equation x2(3x + 1) = 0 and then divide both sides by x2 and solve 

3x + 1 = 0, thus losing the zero solution. Whenever you divide both sides of an algebraic equation by x or a power of x, you incorrectly lose the zero solution.


Example:  Solve ∣2x + 5∣  > 1 and graph the solution

2x + 5  > 1 Given

Solve the related equation 2x + 5  = 1

To solve absolute value equations, you must solve two cases:

2x + 5 = 1  and   -(2x + 5) = 1 

Solve 2x + 5 = 1

2x = -4 Add –5 to both sides using Addition Property of Equality

x = -4/2 = -2 Division Property of Equality

Solve –(2x + 5) = 1

-2x – 5 = 1 Multiply through by –1 using Distributive Property

-2x = 6 Add 5 to both sides using Addition Property of Equality

x = 6/(-2) = -3 Division Property of Equality

Plot the critical values, x = 0 and x =-1/3. Use open circles since this is >.

Plot the critical values, x = 0 and x =-1/3. Use open circles since this is >.

Test arbitrary values of each interval. You may test the value in the original inequality or it’s simplified form.  

In Interval 1, let x=0 in2x + 5 > 1.

20 + 5  > 1  is TRUE, so Interval 1 is part of the solution.

In Interval 2, let x = -2.5 in 2x + 5 > 1.

2(-2.5) + 5  > 1  is FALSE, so Interval 2 is NOT part of the solution.

In Interval 3, let x=- 4 in 2x + 5 > 1.

2(-4) + 5  > 1  is TRUE, so Interval 3 is part of the solution.

We shade Interval 1 and Interval 3, but do not include the endpoints.

The solution is x > -2 or x < -3. In interval notation, we would write this as

(- , -3) U (-2, ).

Method To Solve Rational Inequalities:

  1. Move all terms to one side of the inequality sign by applying the Addition, Subtraction, Multiplication, and Division Properties of Inequalities. You should have only zero on one side of the inequality sign.
  2. Solve the associated equation using an appropriate method. This solution or solutions will make up the set of critical values.  At these values, sign changes occur in the inequality. 
  3. Find all values that result in Division By Zero. These are also critical values for rational inequalities.
  4. Plot the critical values on a number line. Use closed circles ∙ for ≤ and ≥ unless the value results in division by zero – always use open circles for values resulting in division by zero since this value can not be part of the solution!   Always use open circles ο for < and > inequalities.
  5. Test each interval defined by the critical values. If an interval satisfies the inequality, then it is part of the solution.  If it does not satisfy the inequality, then it is not part of the solution. 

In summary, the only difference between solving a rational inequality and a polynomial inequality is that there are additional critical values that result in division by zero, and you never include these additional values as part of the solution, even if it is a   or inequality. 

Solve the related equations (-2x + 10)/(x – 3) = 0  and x – 3 = 0.

(-2x + 10)/(x – 3) = 0  

-2x + 10 = 0          Clear fractions by multiplying both sides by (x-3)

-2x = -10         Add –10 to both sides using Addition Property of Equality

x = -10/(-2) = 5         Division Property of Equality

x – 3 = 0

x = 3         Addition Property of Equality

Plot the critical numbers.  Use a closed circle for x=5 but an open circle for x=3.

In Interval 1, we let x = 6. This results in (-2 6 + 10)/(6 – 3) 0  or 

–2/3 0 which is TRUE.  So Interval 1 is part of the solution.

In Interval 2, we let x = 4. This results in (-2 4 + 10)/(4 – 3) 0  or 

2/1 0 which is FALSE.  So Interval 2 is NOT part of the solution.

In Interval 3, we let x = 2. This results in (-2 2 + 10)/(2 – 3) 0  or 

6/(-1) 0 which is TRUE.  So Interval 3 is part of the solution.

This results in a graph of

with solution x < 3 or x 5.  Interval notation is (- , 3) U [5, ).

Absolute Value Inequalities 

In the previous section we solved equations that contained absolute values. In this section we want to look at inequalities that contain absolute values. We will need to examine two separate cases.

Inequalities Involving < and 

As we did with equations let’s start off by looking at a fairly simple case.

This says that no matter what p is it must have a distance of no more than 4 from the origin. This means that p must be somewhere in the range,

We could have a similar inequality with the < and get a similar result.

In general, we have the following formulas to use here,

Example: 

There really isn’t much to do other than plug into the formula. As with equations pp simply represents whatever is inside the absolute value bars. So, with this first one we have,

Now, this is nothing more than a fairly simple double inequality to solve so let’s do that.

The interval notation for this solution is 

Inequalities Involving > and 

Once again let’s start off with a simple number example.

This says that whatever p is it must be at least a distance of 4 from the origin and so p must be in one of the following two ranges,

These solutions must be written as two inequalities.

Here is the general formula for these.

Example:

Again, p represents the quantity inside the absolute value bars so all we need to do here is plug into the formula and then solve the two linear inequalities.

The interval notation for these are

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