Over Lesson 2–4

A. –4

B. –1

C. 4

D. 13

Solve 8y + 3 = 5y + 15.

Over Lesson 2–4

A. 50

B. 25

C. 2

D. all numbers

Solve 5(x + 3) + 2 = 5x + 17.

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 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.

You solved equations with the variable on each side.

• Evaluate absolute value expressions.

• Solve absolute value equations.

Expressions with Absolute Value

Evaluate |a – 7| + 15 if a = 5.

|a – 7| + 15 = |5 – 7| + 15 Replace a with 5.

= |–2| + 15 5 – 7 = –2

= 2 + 15 |–2| = 2

= 17 Simplify.

Answer: 17

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.

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.

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 1 Case 2t – 1 = 5 t – 1 = –5

t – 1 + 1 = 5 + 1 Add 1 to each side.

t – 1 + 1 = –5 + 1

t = 6 Simplify. t = –4

Solve an Absolute Value Equation

Solve Absolute Value Equations

A. Solve |2x – 1| = 7. Then graph the solution set.

|2x – 1| = 7 Original equation

Case 1 Case 2

2x – 1 = 7 2x – 1 = –7

2x – 1 + 1 = 7 + 1 Add 1 to each side. 2x – 1 + 1 = –7 + 1

2x = 8 Simplify. 2x = –6Divide each side by 2.

x = 4 Simplify. x = –3

Solve Absolute Value Equations

Answer: {–3, 4}

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

A. Solve |2x + 3| = 5. Graph the solution set.

A. {1, –4}

B. {1, 4}

C. {–1, –4}

D. {–1, 4}

B. Solve |x – 3| = –5.

A. {8, –2}

B. {–8, 2}

C. {8, 2}

D.