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

2-4 Solving Equations with Variables on Both Sides

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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


MA.912.A.3.1 Solve linear equations in one variable that include simplifying algebraic expressions.Also MA.912.A.3.5.

Sunshine State Standards


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

Objective


Vocabulary

identity


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 undo the subtraction.


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 To isolate b, subtract 2 from both sides.


Solve 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.


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


Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

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+ 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.


Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Additional Example 2: Continued

40 = 12a

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


Solve .


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


Solve 3x + 15 – 9 = 2(x + 2).


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.

Subtract.


An identity is an equation that is always true, no matter what value is substituted for the variable. The solutions of an identity are all real numbers. Some equations are always false. They have no solutions.


Solve 10 – 5x + 1 = 7x + 11 – 12x.

Additional Example 3A: Infinitely Many Solutions or No Solutions

Add 5x to both sides.

Identify like terms.10 – 5x + 1 = 7x + 11 – 12x

11 – 5x = 11 – 5x

11 = 11

+ 5x + 5x True statement.

Combine like terms on the left and the right.

10 – 5x + 1 = 7x + 11 – 12x

The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.


Solve 12x – 3 + x = 5x – 4 + 8x.

Additional Example 3B: Infinitely Many Solutions or No Solutions

Subtract 13x from both sides.

Identify like terms.12x – 3 + x = 5x – 4 + 8x

13x – 3 = 13x – 4

–3 = –4

–13x –13x False statement.


12x – 3 + x = 5x – 4 + 8x

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


Solve 4y + 7 – y = 10 + 3y.


Subtract 3y from both sides.

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

3y + 7 = 3y + 10

7 = 10

–3y –3y False statement.


4y + 7 – y = 10 + 3y

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


Solve 2c + 7 + c = –14 + 3c + 21.


Subtract 3c both sides.

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

3c + 7 = 3c + 7

7 = 7

–3c –3c True statement.


2c + 7 + c = –14 + 3c + 21

The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. 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


Example 4 Continued

Let b represent bulbs, 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 + 44b = 96 + 32b

60 + 44b = 96 + 32b– 32b – 32b

To collect the variable terms on one side, subtract 32b from both sides.

60 + 12b = 96


Additional Example 4 Continued

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

60 + 12b = 96–60 – 60

12b = 36Since b is multiplied by 12,

divide both sides by 12 to undo the multiplication.

b = 3


Additional Example 4 Continued

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 b = 3:

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

96 + 32b = 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.


Standard Lesson Quiz

Lesson Quizzes

Lesson Quiz for Student Response Systems


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

1

20 hours



1. Solve the equation.

A. a = 1

D. a = 6

C. a = 3

B. a = 2

9a + 3 = 6a + 6




A. p = 2

D. p = 5

C. p = 4

B. p = 3

7(3p − 4) = 4p + 6




A.

B.

C.

D.

5 − 2(3z + 2) = –4(1 − 2z)




A.

B.

C.

D. None of the above

6(s + 2) − 5s = 16 + s − 4

All real numbers

No solution

(0,0)




A.

B.

C.

D.

a = 2

a = 5

a = 4

a = 3



6. A moving company charges a $2,000 fee

plus a $100 per hour. Another moving

company charges an $1800 fee plus $150

per hour. How long is a job for which the

costs of the two companies is the same?

A. 2 h

B. 3 h

C. 4 h

D. 5 h

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

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