Objective: solve equations using addition and subtraction Chapter 1 Algebra Toolbox 1-3 Solving Equations by Adding or Subtracting Objective: solve equations using addition and subtraction

Agenda Warm Up Problem of the Day Lesson Presentation Practice Activity Quiz

Warm Up Write an algebraic expression for each word phrase. 1. a number x decreased by 9 2. 5 times the sum of p and 6 3. 2 plus the product of 8 and n 4. the quotient of 4 and a number c x  9 5(p + 6) 2 + 8n 4 c

Problem of the Day Janie’s horse refused to do 5 jumps today and cleared 14 jumps. Yesterday, the horse cleared 9 more jumps than today. He won 3 first place ribbons. How many jumps did the horse clear in the two-day jumping event? 37

Learn to solve equations using addition and subtraction.

Vocabulary equation solve solution inverse operation isolate the variable Addition Property of Equality Subtraction Property of Equality

An equation uses an equal sign to show that two expressions are equal An equation uses an equal sign to show that two expressions are equal. All of these are equations. 100 2 = 50 3 + 8 = 11 r + 6 = 14 24 = x – 7 To solve an equation, find the value of the variable that makes the equation true. This value of the variable is called the solution of the equation.

Determining If a Number is a Solution of an Equation Example 1: Determining If a Number is a Solution of an Equation Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. x + 8 = 15 ? 5 + 8 = 15 ? Substitute 5 for x. 13= 15 ?  So 5 is not solution.

Determine which value of x is a solution of the equation. Example 1 Continued Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. x + 8 = 15 ? 7 + 8 = 15 ? Substitute 7 for x. 15= 15 ?  So 7 is a solution.

Determine which value of x is a solution of the equation. Example 1 Continued Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. x + 8 = 15 ? 23 + 8 = 15 ? Substitute 23 for x. 31= 15 ?  So 23 is not a solution.

Addition and subtraction are inverse operations, which means they “undo” each other. To solve an equation, use inverse operations to isolate the variable. This means getting the variable alone on one side of the equal sign.

ADDITION PROPERTY OF EQUALITY To solve a subtraction equation, like y  15 = 7, you would use the Addition Property of Equality. ADDITION PROPERTY OF EQUALITY Words Numbers Algebra You can add the same number to both sides of an equation, and the statement will still be true. 2 + 3 = 5 x = y + 4 x = y + z 2 + 7 = 9

SUBTRACTION PROPERTY OF EQUALITY There is a similar property for solving addition equations, like x + 9 = 11. It is called the Subtraction Property of Equality. SUBTRACTION PROPERTY OF EQUALITY Words Numbers Algebra You can subtract the same number from both sides of an equation, and the statement will still be true. 4 + 7 = 11 x = y  3 x = y  z 4 + 4 = 8

Solving Equations Using Addition & Subtraction Properties Example 2A: Solving Equations Using Addition & Subtraction Properties Solve. A. 10 + n = 18 10 + n = 18 –10 –10 Subtract 10 from both sides. 0 + n = 8 n = 8 Identity Property of Zero: 0 + n = n. Check 10 + n = 18 ? 10 + 8 = 18 18 = 18 ? 

Solving Equations Using Addition & Subtraction Properties Example 2B: Solving Equations Using Addition & Subtraction Properties Solve. B. p – 8 = 9 p – 8 = 9 + 8 + 8 Add 8 to both sides. p + 0 = 17 p = 17 Identity Property of Zero: p + 0 = p. Check p – 8 = 9 ? 17 – 8 = 9 9 = 9 ? 

Solving Equations Using Addition & Subtraction Properties Example 2C: Solving Equations Using Addition & Subtraction Properties Solve. C. 22 = y – 11 22 = y – 11 + 11 + 11 Add 11 to both sides. 33 = y + 0 33 = y Identity Property of Zero: y + 0 = 0. Check 22 = y – 11 ? 22 = 33 – 11 22 = 22 ? 

= = + + Example 3A Solve: x 34 16,550 x + 34 = 16,550 –34 – 34 A. Jan took a 34-mile trip in her car, and the odometer showed 16,550 miles at the end of the trip. What was the original odometer reading? odometer reading at the beginning of the trip miles traveled + = odometer reading at the end of the trip + = Solve: x 34 16,550 x + 34 = 16,550 –34 – 34 Subtract 34 from both sides. x + 0 = 16,516 x = 16,516 The original odometer reading was 16,516 miles.

= = + + Example 3B Solve: 895 n 1125 895 + n = 1125 –895 – 895 B. From 1980 to 2000, the population of a town increased from 895 residents to 1125 residents. What was the increase in population during that 20-year period? initial population increase in population + = population after increase + = Solve: 895 n 1125 895 + n = 1125 –895 – 895 Subtract 895 from both sides. 0 + n = 230 n = 230 The increase in population was 230.

Lesson Quiz Determine which value of x is a solution of the equation. 1. x + 9 = 17; x = 6, 8, or 26 2. x – 3 = 18; x = 15, 18, or 21 Solve. 3. a + 4 = 22 4. n – 6 = 39 5. The price of your favorite cereal is now $4.25. In prior weeks the price was $3.69. Write and solve an equation to find n, the increase in the price of the cereal. 8 21 a = 18 n = 45 3.69 + n = 4.25; $0.56

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