holt algebra 1 2-1 solving equations by adding or subtracting 2-1 solving one step equations using...

Holt Algebra 1

2-1Solving Equations by Adding or Subtracting2-1 Solving One Step Equations using

inverse operations.

Holt Algebra 1

Lesson PresentationLesson Presentation

2-2

Holt Algebra 1

2-1Solving Equations by Adding or Subtracting

Solve one-step equations in one variable by using addition or subtraction.

Objective

Solve one-step equations in one variable by using multiplication or division.

Holt Algebra 1


An equation is a mathematical statement that two expressions are equal.

A solution of an equation is a value of the variable that makes the equation true.

To find solutions, isolate the variable. A variable is isolated when it appears by itself on one side of an equation, and not at all on the other side.

Holt Algebra 1


Inverse Operations

Operation Inverse Operation

Addition Subtraction

Subtraction Addition

Isolate a variable by using inverse operations which "undo" operations on the variable.

An equation is like a balanced scale. To keep the balance, perform the same operation on both sides.

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WORDS

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

NUMBERS

3 = 3 3 + 2 = 3 + 2 5 = 5

ALGEBRA a = b a + c = b + c

Properties of Equality

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WORDS

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

NUMBERS

7 = 7 7 – 5 = 7 – 5 2 = 2

ALGEBRA a = b a – c = b – c


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Solve the equation. Check your answer.

Example 1A: Solving Equations by Using Addition

Since 8 is subtracted from y, add 8 to both sides to undo the subtraction.

y – 8 = 24 + 8 + 8

y = 32

Check y – 8 = 24

32 – 8 2424 24

To check your solution, substitute 32 for y in the original equation.

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Example 1B: Solving Equations by Using Addition

= z34

To check your solution, substitute for z in the original equation.

34

+ 716

+ 716

= z – 716

516

Check= z – 7

16 516

34

516

716

–

516

516

Since is subtracted from z, add to

both sides to undo the subtraction.

716

716

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Check It Out! Example 1a

Since 3.2 is subtracted from n, add 3.2 to both sides to undo the subtraction.

n – 3.2 = 5.6

+ 3.2 + 3.2

n = 8.8

Check n – 3.2 = 5.6

8.8 – 3.2 5.65.6 5.6

To check your solution, substitute 8.8 for n in the original equation.

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Check It Out! Example 1c

Since 9 is subtracted from m, add 9 to both sides to undo the subtraction.

16 = m – 9 + 9 + 9

25 = m

Check 16 = m – 9

16 25 – 916 16

To check your solution, substitute 25 for m in the original equation.

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Remember that subtracting is the same as adding the opposite. When solving equations, you will sometimes find it easier to add an opposite to both sides instead of subtracting.

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Example 3: Solving Equations by Adding the Opposite

p = 311

+ 511

+ 511

Solve – + p = – . 211

511

511

Since – is added to p, add

to both sides.

511

Check + p = – 211

511

–

2 511 11

– – 311

+

211

– 211

–

To check your solution, substitute for p in the original equation.

311

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Solve –2.3 + m = 7. Check your answer.


Since –2.3 is added to m, add 2.3 to both sides.

–2.3 + m = 7+2.3 + 2.3

m = 9.3

Check –2.3 + m = 7 –2.3 + 9.3 7

7 7

To check your solution, substitute 9.3 for m in the original equation.

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Check It Out! Example 3b

z = 2

+ 3 4

+ 3 4

Solve – + z = . Check your answer. 5 4

3 4

3 4

Since – is added to z, add

to both sides.

3 4

Check + z = 5 4

3 4

–

– 5 3 4 4

+ 2

5 4

5 4

To check your solution, substitute 2 for z in the original equation.

– + z =

5 4

3 4

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Over 20 years, the population of a town decreased by 275 people to a population of 850. Write and solve an equation to find the original population.

Example 4: Application

Write an equation to represent the relationship.

+ 275 + 275

p =1125

p – d = c

original population minus

current population

decrease in

populationis

p – 275 = 850 Since 275 is subtracted from p, add 275 to both sides to undo the subtraction.

p – d = c

The original population was 1125 people.

Holt Algebra 1


A person's maximum heart rate is the highest rate, in beats per minute, that the person's heart should reach. One method to estimate maximum heart rate states that your age added to your maximum heart rate is 220. Using this method, write and solve an equation to find a person's age if the person's maximum heart rate is 185 beats per minute.

Check It Out! Example 4

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Check It Out! Example 4 Continued

a + r = 220

ageadded to

220maximum heart rate is


– 185 – 185

a = 35

a + 185 = 220 Substitute 185 for r. Since 185 is added to a, subtract 185 from both sides to undo the addition.

a + r = 220

A person whose maximum heart rate is 185 beats per minute would be 35 years old.

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Inverse Operations

Operation Inverse Operation

Multiplication Division

Division Multiplication

Solving an equation that contains multiplication or division is similar to solving an equation that contains addition or subtraction. Use inverse operations to undo the operations on the variable.

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WORDS

Multiplication Property of EqualityYou can multiply both sides of an equation by the same number, and the statement will still be true.

NUMBERS

6 = 6 6(3) = 6(3) 18 = 18

ALGEBRA a = b ac = bc


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Properties of EqualityDivision Property of EqualityYou can divide both sides of an equation by the same nonzero number, and the statement will still be true.

WORDS

a = b (c ≠ 0)

8 = 8

2 = 2

ALGEBRA

NUMBERS 84

84

=

ac

ac=

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Solve the equation.

Example 1A: Solving Equations by Using Multiplication

Since j is divided by 3, multiply both sides by 3 to undo the division.–24 = j

–8 –8

To check your solution, substitute –24 for j in the original equation.

–8 =j3

–8 –243

Check –8 =j3

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Solve the equation.

Example 1B: Solving Equations by Using Multiplication

Since n is divided by 6, multiply both sides by 6 to undo the division.n = 16.8

2.8 2.8

To check your solution, substitute 16.8 for n in the original equation.

= 2.8n6

2.8 16.86

Check = 2.8n6

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Example 2B: Solving Equations by Using Division

Since v is multiplied by –6, divide both sides by –6 to undo the multiplication.

0.8 = v

–4.8 –4.8

To check your solution, substitute 0.8 for v in the original equation.

–4.8 = –6v

–4.8 –6(0.8)

Check –4.8 = –6v

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Since c is multiplied by 4, divide both sides by 4 to undo the multiplication.

4 = c

16 16

To check your solution, substitute 4 for c in the original equation.

16 = 4c

16 4(4)

Check 16 = 4c

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Check It Out! Example 2b

Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication.y = –20

–10 –10

To check your solution, substitute –20 for y in the original equation.

0.5y = –10

0.5(–20) –10

Check 0.5y = –10

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Remember that dividing is the same as multiplying by the reciprocal. When solving equations, you will sometimes find it easier to multiply by a reciprocal instead of dividing. This is often true when an equation contains fractions.

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Solve the equation.

Example 3A: Solving Equations That Contain Fractions

w = 24

20 20

To check your solution, substitute 24 for w in the original equation.

w = 2056

Check w = 2056

The reciprocal of is . Since w is

multiplied by , multiply both sides

by .

56

65

566

5

20

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Solve the equation.

Example 3B: Solving Equations That Contain Fractions

= z3

16

To check your solution,

substitute for z in the

original equation.

32

= z32

18

The reciprocal of is 8. Since z is


by 8.

181

8

Check18

316

= z

316

316

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Solve the equation. Check your answer.Check It Out! Example 3a

– = b14

To check your solution,

substitute – for b in the

original equation.

54

15

The reciprocal of is 5. Since b is


by 5.

151

5

= b54–

= b5 4Check11–

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Example 4: Application


Ciro puts of the money he earns from mowing lawns into a college education fund. This year Ciro added $285 to his college education fund. Write and solve an equation to find how much money Ciro earned mowing lawns this year.

14

one-fourth times earnings equals college fund

m = $1140

Substitute 285 for c. Since m is divided by 4, multiply both sides by 4 to undo the division.

Ciro earned $1140 mowing lawns.

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Check it Out! Example 4


The distance in miles from the airport that a plane should begin descending, divided by 3, equals the plane's height above the ground in thousands of feet. A plane began descending 45 miles from the airport. Use the equation to find how high the plane was flying when the descent began.

Distance divided by 3 equals height in thousands of feet

15 = h

Substitute 45 for d.

The plane was flying at 15,000 ft when the descent began.

Holt Algebra 1


Homework

• Problems to try before you come to class.

• Page80 #22 to 48 even• Page87-88 #22-42

even

• Don’t forget to bring your textbook to class.

holt algebra 1 2-1 solving equations by adding or subtracting 2-1 solving one step equations using...

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