1-3 solving addition and subtraction equations pre-algebra 1-3 solving addition and subtraction...

1-3 Solving Addition and Subtraction Equations

Pre-Algebra


Pre-Algebra

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation


Pre-Algebra

Warm UpWrite an algebraic expression for each word phrase.

1. a number x decreased by 92. 5 times the sum of p and 63. 2 plus the product of 8 and n4. the quotient of 4 and a number c

x 95(p + 6)

2 + 8n4c


Pre-Algebra


Pre-Algebra

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


Pre-Algebra

Learn to solve equations using addition and subtraction.


Pre-Algebra

Vocabularyequationsolvesolutioninverse operationisolate the variableAddition Property of EqualitySubtraction Property of Equality


Pre-Algebra

An equation uses an equal sign to show that two expressions are equal. All of these are equations.

3 + 8 = 11 r + 6 = 14 24 = x – 7 1002

= 50

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.


Pre-Algebra

Determine which value of x is a solution of the equation.

x + 8 = 15; x = 5, 7, or 23

Additional Example 1: Determining Whether a Number is a Solution of an Equation

Substitute each value for x in the equation.

Substitute 5 for x.13= 15 ?

So 5 is not solution.

x + 8 = 15?

5 + 8 = 15?


Pre-Algebra

Determine which value of x is a solution of the equation.x + 8 = 15; x = 5, 7, or 23

Additional Example 1 Continued



So 7 is a solution.

x + 8 = 15?

7 + 8 = 15?


Pre-Algebra

Determine which value of x is a solution of the equation.x + 8 = 15; x = 5, 7, or 23

Additional Example 1 Continued



So 23 is not a solution.

x + 8 = 15?

23 + 8 = 15 ?


Pre-Algebra

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.


Pre-Algebra

To solve a subtraction equation, like y 15 = 7, you would use the Addition Property of Equality.

You can add the same number to both sides of an equation, and the statement will still be true.

2 + 3 = 5+ 4 + 4

2 + 7 = 9

x = yx = y+ z + z

ADDITION PROPERTY OF EQUALITY

Words Numbers Algebra


Pre-Algebra

There is a similar property for solving addition equations, like x + 9 = 11. It is called the Subtraction Property of Equality.

You can subtract the same number from both sides of an equation, and the statement will still be true.

4 + 7 = 11 3 3

4 + 4 = 8

x = yx = y z z

SUBTRACTION PROPERTY OF EQUALITY

Words Numbers Algebra


Pre-Algebra

Solve.

Additional Example 2A: Solving Equations Using Addition and Subtraction Properties

Subtract 10 from both sides.

A. 10 + n = 1810 + n = 18

–10 –10

0 + n = 8 n = 8 Identity Property of Zero: 0 + n = n.

Check

10 + n = 18?

10 + 8 = 18

18 = 18?


Pre-Algebra

Solve.

Additional Example 2B: Solving Equations Using Addition and Subtraction Properties

Add 8 to both sides.

B. p – 8 = 9p – 8 = 9

+ 8 + 8

p + 0 = 17 p = 17 Identity Property of Zero: p + 0 = p.

Checkp – 8 = 9

? 17 – 8 = 9

9 = 9?


Pre-Algebra

Solve.

Additional Example 2C: Solving Equations Using Addition and Subtraction Properties

Add 11 to both sides.

C. 22 = y – 1122 = y – 11

+ 11 + 11

33 = y + 0 33 = y Identity Property of Zero: y + 0 = 0.

Check22 = y – 11

? 22 = 33 – 11

22 = 22?


Pre-Algebra

Additional Example 3A

+ =+ =34 16,550

x + 0 = 16,516

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?


x + 34 = 16,550

The original odometer reading was 16,516 miles.

odometer reading at the beginning of the

tripmiles traveled

x

–34 – 34

x = 16,516

Solve:

odometer reading at the end of the trip


Pre-Algebra

Additional Example 3B

+ =+ =n 1125

0 + n = 230

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?


895 + n = 1125

The increase in population was 230.

initial population increase in population

895

–895 – 895

n = 230

Solve:

population after increase


Pre-Algebra

Lesson QuizDetermine 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.

821

a = 18

n = 45

3.69 + n = 4.25; $0.56

1-3 solving addition and subtraction equations pre-algebra 1-3 solving addition and subtraction...

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