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Over Lesson 2–4 5-Minute Check 1 A.–4 B.–1 C.4 D.13 Solve 8y + 3 = 5y + 15.

Over Lesson 2–4 5-Minute Check 4 A.50 B.25 C.2 D.all numbers Solve 5(x + 3) + 2 = 5x + 17.

CCSS Content Standards A.REI.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. A.REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. Mathematical Practices 3 Construct viable arguments and critique the reasoning of others. 7 Look for and make use of structure. Common Core State Standards © Copyright National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.

Then/Now You solved equations with the variable on each side. Evaluate absolute value expressions. Solve absolute value equations.

Example 1 Expressions with Absolute Value Evaluate |a – 7| + 15 if a = 5. |a – 7| + 15= |5 – 7| + 15Replace a with 5. = |–2| – 7 = –2 = |–2| = 2 = 17Simplify. Answer: 17

Concept

Example 3 Solve an Absolute Value Equation WEATHER The average January temperature in a northern Canadian city is 1°F. The actual January temperature for that city may be about 5°F warmer or colder. Write and solve an equation to find the maximum and minimum temperatures. Method 1 Graphing |t – 1| = 5 means that the distance between t and 1 is 5 units. To find t on the number line, start at 1 and move 5 units in either direction.

Example 3 Solve an Absolute Value Equation The solution set is {–4, 6}. The distance from 1 to 6 is 5 units. The distance from 1 to –4 is 5 units.

Example 3 Method 2 Compound Sentence Write |t – 1| = 5 as t – 1 = 5 or t – 1 = –5. Answer: The solution set is {–4, 6}. The maximum and minimum temperatures are –4°F and 6°F. Case 1Case 2 t – 1 = 5t – 1 = –5 t – = 5 + 1Add 1 to each side. t – = –5 + 1 t = 6Simplify. t = –4 Solve an Absolute Value Equation

Example 2 Solve Absolute Value Equations A. Solve |2x – 1| = 7. Then graph the solution set. |2x – 1| = 7Original equation Case 1 Case 2 2x – 1= 7 2x – 1= –7 2x – = Add 1 to each side. 2x – = – x= 8 Simplify. 2x= –6 Divide each side by 2. x= 4 Simplify. x= –3

Example 2 Solve Absolute Value Equations Answer: {–3, 4}

Example 2 Solve Absolute Value Equations B. Solve |p + 6| = –5. Then graph the solution set. |p + 6| = –5 means the distance between p and 6 is –5. Since distance cannot be negative, the solution is the empty set Ø. Answer: Ø

Example 2 A. Solve |2x + 3| = 5. Graph the solution set. A.{1, –4} B.{1, 4} C.{–1, –4} D.{–1, 4}

Example 2 B. Solve |x – 3| = –5. A.{8, –2} B.{–8, 2} C.{8, 2} D.