Ultimate guide to linear inequalities

Ultimate Guide

To

Solving

Linear Inequalities

Solving Linear Inequalities

1.4 Sets, Inequalities, and Interval Notation

1.5 Intersections, Unions, and Compound Inequalities

1.6 Absolute-Value Equations and Inequalities

OBJECTIVES


a Determine whether a given number is a solution of aninequality.

b Write interval notation for the solution set or the graphof an inequality.

c Solve an inequality using the addition principle and themultiplication principle and then graph the inequality.

d Solve applied problems by translating to inequalities.


Inequality

An inequality is a sentence containing


Solution of an Inequality

Any replacement or value for the variable that makes an inequality true is called a solution of the inequality. The set of all solutions is called the solution set. When all the solutions of an inequality have been found, we say that we have solved the inequality.

EXAMPLE


aDetermine whether a given number is a solution of aninequality.

1 Determine whether the given number is a solution of the inequality.

EXAMPLE Solution



1

We substitute 5 for x and get 5 + 3 < 6, or 8 < 6, a false sentence. Therefore, 5 is not a solution.

EXAMPLE



3 Determine whether the given number is a solution of the inequality.

EXAMPLE Solution



3

We substitute –3 for x and get or a true sentence. Therefore, –3 is a solution.


bWrite interval notation for the solution set or the graph of an inequality.

The graph of an inequality is a drawing that represents its solutions.

EXAMPLE



4 Graph on the number line.

EXAMPLE Solution



4

The solutions are all real numbers less than 4, so we shade all numbers less than 4 on the number line. To indicate that 4 is not a solution, we use a right parenthesis “)” at 4.





EXAMPLE



Write interval notation for the given set or graph.

EXAMPLE Solution




cSolve an inequality using the addition principle and themultiplication principle and then graph the inequality.

Two inequalities are equivalent if they have the same solution set.


The Addition Principle for Inequalities

EXAMPLE



10 Solve and graph.

EXAMPLE Solution



10

We used the addition principle to show that the inequalities x + 5 > 1 and x > –4 are equivalent. The solution set is and consists of an infinite number of solutions. We cannot possibly check them all.

EXAMPLE Solution



10Instead, we can perform a partial check by substituting one member of the solution set (here we use –1) into the original inequality:

EXAMPLE Solution

Since 4 > 1 is true, we have a partial check. The solution set is or The graph is as follows:



10

EXAMPLE



11 Solve and graph.

EXAMPLE Solution



11

EXAMPLE Solution



11

The inequalities and have the same meaning and the same solutions. The solution set is or more commonly, Using interval notation, we write that the solution set is The graph is as follows:


The Multiplication Principle for Inequalities

For any real numbers a and b, and any positive number c:

For any real numbers a and b, and any negative number c:

Similar statements hold for


The multiplication principle tells us that when we multiply or divide on both sides of an inequality by a negative number, we must reverse the inequality symbol to obtain an equivalent inequality.

EXAMPLE



13 Solve and graph.

EXAMPLE Solution



13

EXAMPLE Solution



13

EXAMPLE



15 Solve.

EXAMPLE Solution



15

EXAMPLE Solution



15






Translating “At Least” and “At Most”

EXAMPLE



16 Cost of Higher Education.

The equation C = 126t + 1293 can be used to estimate the average cost of tuition and fees at two-year public institutions of higher education, where t is the number of years after 2000. Determine, in terms of an inequality, the years for which the cost will be more than $3000.

EXAMPLE Solution



16

1. Familiarize. We already have a formula. To become more familiar with it, we might make a substitution for t. Suppose we want to know the cost 15 yr after 2000, or in 2015. We substitute 15 for t:

C = 126(15) + 1293 = $3183.

EXAMPLE Solution



16

We see that in 2015, the cost of tuition and fees at two-year public institutions will be more than $3000. To find all the years in which the cost exceeds $3000, we could make other guesses less than 15, but it is more efficient to proceed to the next step.

EXAMPLE Solution



16

2. Translate. The cost C is to be more than $3000. Thus we have C > 3000. We replace C with 126t + 1293 to find the values of t that are solutions of the inequality:

126t + 1293 > 3000.

EXAMPLE Solution



16

3. Solve. We solve the inequality:

EXAMPLE Solution



16

4. Check. A partial check is to substitute a value for t greater than 13.55. We did that in the Familiarize step and found that the cost was more than $3000.

5. State. The average cost of tuition and fees at two-year public institutions of higher education will be more than $3000 for years more than 13.55 yr after 2000, so we have

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Ultimate guide to linear inequalities

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