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Inequalities in Algebra
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Inequalities in Algebra

Introduction:

An inequality is similar to an equation. Here two expressions are separated by a symbol that indicates how one expression is related to the other. Where as In an equation such as 8x = 56, the sign (=) indicates that the expressions are equivalent, in case of an inequality, such as 8x > 56, the > sign indicates that the L.H.S is larger than the R.H.S. The symbol (<) is used in inequalities, which means "less than".

To solve the inequality 8x > 56, you can follow the same rules that for equations. In this case, divide both sides by 8 so that x > 7.

Remember: In this example, x is a value and it is always larger than 7. It'll never be equal to or less than 7.

Solving Inequalities:
An inequality is the result of replacing the = sign in an equation with <, >, , or . E.g. 6x - 4 < 14 is a linear inequality. If < was replaced by = sign, it would be a linear equation. In case of polynomial inequalities, it involves polynomials of degree two or more, e.g. 2x3 -x > 0, can be referred to as polynomial inequalities. In case the degree is exactly 2, it is called quadratic inequality. Similarly rational inequalities are those, where rational expressions are involved.

Notice that solving inequalities is about: sign changes. Identify all the points at which there are sign changes (these points are known as critical values). Then determine which, if any, of the intervals bound by these critical values result in a solution. The set of all such points contained in the solution intervals, will be the solution to inequality.

Solving Linear, Polynomial, or Absolute Value Inequalities:

Step 1: Move all the terms to one side of the inequality sign by using the Addition, Subtraction, Multiplication, or Division Properties of Inequalities. Make sure that you have only zero on one side of the inequality sign.
Step 2: Solve the associated equation using a suitable and applicable method. This solution will yield the set of critical values. At these values, sign changes occur in the inequality.
Step 3: In case, you are required to graph the solution, plot the critical values on a number line, using closed circles for and inequalities, and open circles o for < and > inequalities.
Step 4: Then you should test each interval defined by the critical values. If an interval satisfies the inequality, it is part of the solution. If it does not satisfy the inequality, then it is not.

Solving Rational Inequalities:

Step 1: Move all the terms to one side of the inequality sign by using the Addition, Subtraction, Multiplication, or Division Properties of Inequalities. Make sure that you have only zero on one side of the inequality sign.
Step 2: Solve the associated equation using a suitable and applicable method. This solution will yield the set of critical values. At these values, sign changes occur in the inequality.
Step 3: Find all values that result in Division by Zero. These are also critical values for rational inequalities.
Step 4: In case you are required to graph the solution, plot the critical values on a number line, using 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 o for < and > inequalities.
Step 5: Then you should 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.

Remember: The main difference between solving a polynomial inequality and rational inequality is that there are additional critical values that result in division by zero, and these additional values are not included as part of the solution, even if it is a or inequality.
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The following example will illustrate the steps, how to solve the inequalities:
Example: Solve 5x + 4(x + 1) 5x - 1
Solution: 5x + 4(x + 1) 5x - 1       Given
5x + 4x + 4 5x - 1        Distributive Property
9x + 4 5x - 1       Combine Like Terms
4x + 5 0 Subtract 5x from both sides, add 1 to both sides using
Addition and Subtraction Properties of Inequality
Now, solve 4x + 5 = 0
4x = -5         Addition Property of Equality
x = - 5/4          Division Property of Equality
The solution is x - 5/4, or in interval notation, (- , - 5/4].

You can test the solution by substituting the arbitary values
let x=1 in 4x + 5 0.
4(1) + 5 0 is FALSE.

Now, let x = -2 in 4x + 5 0.
4(-2) + 5 0 is TRUE.

Example: Solve 2x3 + 4x2 > 3x2
Solution: 2x3 + 4x2 > 3x2      Given
2x3 + x2 > 0       Subtraction Property of Inequalities
Now, solve 2x3 + x2 = 0
x2(2x + 1) = 0     Distributive Property
x2 = 0 or 2x + 1 = 0      Zero Product Law
Solve x2 = 0
x=0 = 0     Extract Square Roots
Solve 2x + 1 = 0
2x = -1 Add -1 to using Addition Property of Equality
x = -1/2        Division Property of Equality
The solution is x > 0 or -1/2 < x < 0.
The interval notation of this solution is (0 , ) U (-1/2, 0).

You can test the solution by substituting the arbitary values
let x=1 in x2(2x + 1) > 0.
12(2(1) + 1) > 0 is TRUE.

let x=-1/4 in x2(2x + 1) > 0.
(-1/4)2(2(-1/4) + 1) > 0 is TRUE, since simplified, we get (1/16)(1/2) > 0.

let x=-1 in x2(2x + 1) > 0.
(-1)2(2(-1) + 1) > 0 is FALSE, since simplified, we get (1)(-1) > 0
 
   
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