Algeba 1. 9.1 Solving equations review

Holt Algebra 1

2-4Solving Equations with Variables on Both Sides

BellringerSimplify.

1. 15x – 4x

2. 5(x – 7)

Solve.

3. x + 7 = 18

4. 5a – 6 = 44

11x5x - 35

x = 11

x = 10

Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2

Holt Algebra 1

2-4Solving Equations with Variables on Both Sides2-4 Solving Equations with

Variables on Both Sides

Holt Algebra 1

•BellringerBellringer

•Review for QuizReview for Quiz

•Wall reviewWall review

•Game (If we have time)Game (If we have time)

Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2, CLE 3102.3.2

Holt Algebra 1


Solve equations in one variable that contain variable terms on both sides.


Objective


Holt Algebra 1


Solve the equation. Check your answer.

Review…

16 = m – 8

+ 8 + 824 = m


Holt Algebra 1


Solve 2a + 3 – 8a = 8.

Check It Out! Example 3a

Use the Commutative Property of Addition.

2a + 3 – 8a = 8

2a – 8a + 3 = 8

–6a + 3 = 8 Combine like terms.

Since 3 is added to –6a, subtract 3 from both sides to undo the addition.

– 3 – 3

–6a = 5Since a is multiplied by –6, divide both

sides by –6 to undo the multiplication.

Holt Algebra 1


Solve the equation. Check your answer.

Review….

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.

Holt Algebra 1


Solve 26 = 4a + 10.

Review…

26 = 4a + 10

–10 – 10

16 = 4a

16 = 4a4 4

4 = a


Holt Algebra 1


Solve 5t – 2 = –32.

Review…

5t – 2 = –32

+ 2 + 2

5t = –30

5t = –30 5 5

t = –6


Holt Algebra 1


Solve .

Check it Out! Example 1c

First n is divided by 7. Then 2 is added. Work backward: Subtract 2 from both sides.–2 –2

Since n is divided by 7, multiply both sides by 7 to undo the division.

n = 0

Holt Algebra 1


Solve 7n – 2 = 5n + 6.

Example 1: Solving Equations with Variables on Both Sides

To collect the variable terms on one side, subtract 5n from both sides.

7n – 2 = 5n + 6

–5n –5n

2n – 2 = 6

Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.

2n = 8

+ 2 + 2

n = 4


Holt Algebra 1


Solve 4b + 2 = 3b.

Check It Out! Example 1a

To collect the variable terms on one side, subtract 3b from both sides.

4b + 2 = 3b

–3b –3b

b + 2 = 0

b = –2

– 2 – 2


Holt Algebra 1


Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Example 2: Simplifying Each Side BeforeSolving Equations

Combine like terms.

Distribute –5 to the expression in parentheses.

4 – 6a + 4a = –1 –5(7 – 2a)

4 – 6a + 4a = –1 –5(7) –5(–2a)

4 – 6a + 4a = –1 – 35 + 10a

4 – 2a = –36 + 10a

+36 +36

40 – 2a = 10a+ 2a +2a

40 = 12a

Since –36 is added to 10a, add 36 to both sides.

To collect the variable terms on one side, add 2a to both sides.Standards: SPI 3102.1.3, 3102.2.2,

CLE 3102.3.2, CLE 3102.3.2

Holt Algebra 1


Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Example 2 Continued

40 = 12a

Since a is multiplied by 12, divide both sides by 12.

Standards: SPI 3102.1.3, 3102.2.2, CLE 3102.3.2, CLE 3102.3.2

Holt Algebra 1


Solve 3x + 15 – 9 = 2(x + 2).

Check It Out! Example 2B

Combine like terms.

Distribute 2 to the expression in parentheses.

3x + 15 – 9 = 2(x + 2)

3x + 15 – 9 = 2(x) + 2(2)

3x + 15 – 9 = 2x + 4

3x + 6 = 2x + 4

–2x –2x

x + 6 = 4 – 6 – 6

x = –2

To collect the variable terms on one side, subtract 2x from both sides.

Since 6 is added to x, subtract 6 from both sides to undo the addition.Standards: SPI 3102.1.3, 3102.2.2,

CLE 3102.3.2, CLE 3102.3.2

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Algeba 1. 9.1 Solving equations review