2-4 Solving Equations with Variables on Both Sides Warm Up Warm Up Lesson Presentation Lesson...

2-4 Solving Equations with Variables on Both Sides

Warm UpWarm Up

Lesson Presentation

California Standards

PreviewPreview


Warm UpSimplify.

1. 4x – 10x

2. –7(x – 3)

3.

4. 15 – (x – 2)

Solve.

5. 3x + 2 = 8

6.

–6x–7x + 21

17 – x

2x + 3

2

28


4.0 Students simplify expressions before solving linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12.

5.0 Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step.

California Standards


Vocabulary

identity


To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation.

Helpful HintEquations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms.


Solve 7n – 2 = 5n + 6.

Additional 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

Since 2 is subtracted from 2n, add 2 to both sides.


Solve the equation. Check your answer.

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

4b + 2 = 3b

Since 2 is added to b, subtract 2 from both sides.


Check It Out! Example 1a Continued


4b + 2 = 3b CheckTo check your answer,

substitute –2 for b. 4(–2) + 2 3(–2)

–8 + 2 –6

–6 –6


0.5 + 0.3y = 0.7y – 0.3

Check It Out! Example 1b

To collect the variable terms on one side, subtract 0.3y from both sides.

0.5 + 0.3y = 0.7y – 0.3

–0.3y –0.3y

0.5 = 0.4y – 0.3

0.8 = 0.4y

+0.3 + 0.3

2 = y

Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction.

Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.



Check It Out! Example 1b Continued


0.5 + 0.3y = 0.7y – 0.3

To check your answer, substitute 2 for y.

Check

0.5 + 0.3(2) 0.7(2) – 0.3

1.1 1.1 0.5 + 0.6 1.4 – 0.3


To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.


Solve the equation.

Additional 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

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

4 – 6a + 4a = –1 – 5(7 – 2a)


Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Additional Example 2 Continued

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

40 – 2a = 10a+ 2a +2a

40 = 12a

To collect the variable terms on one side, add 2a to both sides.



Since 1 is subtracted from b, add 1 to both sides.

Distribute to the expression in parentheses.

12

+ 1 + 13 = b – 1

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

12


4 = b


Check It Out! Example 2a Continued


Check

To check your answer, substitute 4 for b.

5 5


3x + 15 – 9 = 2(x + 2)


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.



Check It Out! Example 2b Continued


Check

To check your answer, substitute –2 for x.

3x + 15 – 9 = 2(x + 2)

3(–2) + 15 – 9 2(–2 + 2)

–6 + 15 – 9 2(0)

0 0


An identity is an equation that is always true, no matter what value is substituted for the variable. The solution set of an identity is all real numbers. Some equations are always false. Their solution sets are empty. In other words, their solution sets contain no elements.


Solve the equation.

Additional Example 3A: Infinitely Many Solutions or No Solutions

Identify like terms.

11 – 5x = 11 – 5x Combine like terms on

the left and the right.

The statement 11 – 5x = 11 – 5x is true for all values of x. The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. In other words, all real numbers are solutions.

10 – 5x + 1 = 7x + 11 – 12x

10 – 5x + 1 = 7x + 11 – 12x


Additional Example 3B: Infinitely Many Solutions or No Solutions

Subtract 13x from both sides.

Identify like terms.

13x – 3 = 13x – 4

–3 = –4

–13x –13x

Combine like terms on the left and the right.

The equation 12x – 3 + x = 5x – 4 + 8x is always false. There is no value of x that will make the equation true. There are no solutions.

Solve the equation.

False statement; the solution set is .

12x – 3 + x = 5x – 4 + 8x

12x – 3 + x = 5x – 4 + 8x


The empty set can be written as or {}.

Writing Math


Solve the equation.


Subtract 3y from both sides.

Identify like terms.4y + 7 – y = 10 + 3y

3y + 7 = 3y + 10

7 = 10

–3y –3y False statement; the

solution set is .

4y + 7 – y = 10 + 3y

The equation 4y + 7 – y = 10 + 3y is always false. There is no value of y that will make the equation true. There are no solutions.


Solve the equation.


Identify like terms.2c + 7 + c = –14 + 3c + 21

3c + 7 = 3c + 7 Combine like terms on the left and the right.

2c + 7 + c = –14 + 3c + 21

The statement 3c + 7 = 3c + 7 is true for all values of c. The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. In other words, all real numbers are solutions.


Jon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs will that be?

Additional Example 4: Application

Person Bulbs

Jon 60 bulbs plus 44 bulbs per hour

Sara 96 bulbs plus 32 bulbs per hour



Let h represent hours, and write expressions for the number of bulbs planted.

60 bulbs

plus

44 bulbs each hour

the same

as

96 bulbs

plus

32 bulbs each hour

When is ?

60 + 44h = 96 + 32h

60 + 44h = 96 + 32h To collect the variable terms on one side, subtract 32h from both sides.

60 + 12h = 96

–32h –32h


Since 60 is added to 12h, subtract 60 from both sides.

60 + 12h = 96–60 – 60

12h = 36Since h is multiplied by 12,

divide both sides by 12 to undo the multiplication.

h = 3



After 3 hours, Jon and Sara will have planted the same number of bulbs. To find how many bulbs they will have planted in 3 hours, evaluate either expression for h = 3:

60 + 44h = 60 + 44(3) = 60 + 132 = 192

96 + 32h = 96 + 32(3) = 96 + 96 = 192

After 3 hours, Jon and Sara will each have planted 192 bulbs.



Four times Greg's age, decreased by 3 is equal to 3 times Greg's age, increased by 7. How old is Greg?

Check It Out! Example 4

Let g represent Greg's age, and write expressions for his age.

Four times Greg's

age

decreased by

3is

equal to

three times Greg's

age

increased by

7 .

4g – 3 = 3g + 7


Check It Out! Example 4 Continued

4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides.

g – 3 = 7

–3g –3g

Since 3 is subtracted from g, add 3 to both sides.

+ 3 + 3

g = 10

Greg is 10 years old.


Lesson QuizSolve each equation.

1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4

3. 6 – 7(a + 1) = –3(2 – a)

4. 4(3x + 1) – 7x = 6 + 5x – 2

5.

6. A painting company charges $250 base plus $16 per hour. Another painting company charges $210 base plus $18 per hour. How long is a job for which the two companies costs are the same?

3 8

all real numbers

120 hours

2-4 Solving Equations with Variables on Both Sides Warm Up Warm Up Lesson Presentation Lesson...

Documents

Transcript of 2-4 Solving Equations with Variables on Both Sides Warm Up Warm Up Lesson Presentation Lesson...