Lesson 5 Contents Glencoe McGraw-Hill Mathematics Algebra 2005 Example 1Solve an Absolute Value Equation Example 2Write an Absolute Value Equation Example 3Solve an Absolute Value Inequality (<) Example 4Solve an Absolute Value Inequality (>)

Example 5-1a Method 1 Graphing means that the distance between b and –6 is 5 units. To find b on the number line, start at –6 and move 5 units in either direction. The distance from –6 to –11 is 5 units. The distance from –6 to –1 is 5 units. Answer: The solution set is

Example 5-1a Method 2 Compound Sentence Answer: The solution set is Writeasor Original inequality Subtract 6 from each side. Case 1Case 2 Simplify.

Example 5-1b Answer: {12, –2}

Example 5-2a Write an equation involving the absolute value for the graph. Find the point that is the same distance from –4 as the distance from 6. The midpoint between –4 and 6 is 1. The distance from 1 to –4 is 5 units. The distance from 1 to 6 is 5 units. So, an equation is.

Example 5-2a Check Substitute –4 and 6 into Answer:

Example 5-2b Write an equation involving the absolute value for the graph. Answer:

Example 5-3a Then graph the solution set. Writeasand Original inequality Add 3 to each side. Simplify. Case 1 Case 2 Answer: The solution set is

Example 5-3b Then graph the solution set. Answer:

Example 5-4a Case 1 Case 2 Then graph the solution set. Writeasor Add 3 to each side. Simplify. Original inequality Divide each side by 3. Simplify.

Example 5-4a Answer: The solution set is

Example 5-4b Then graph the solution set. Answer: