1. Preview
Warm Up
California Standards
Lesson Presentation

2. Warm Up Simplify. 1. 4x – 10x 2. –7(x – 3) 3. –6x 4. 15 – (x – 2)
Solve.
5. 3x + 2 = 8 6.
–6x
–7x + 21
2x + 3
17 – x 2 28

3. California
Standards
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.

4. Vocabulary
identity

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

6. Additional Example 1: Solving Equations with Variables on Both Sides
Solve 7n – 2 = 5n + 6.
7n – 2 = 5n + 6
To collect the variable terms on one side, subtract 5n from both sides.
–5n –5n
2n – 2 =
Since 2 is subtracted from 2n, add 2 to both sides.
2n =
Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.
n = 4

7. Check It Out! Example 1a
Solve the equation. Check your answer.
4b + 2 = 3b
4b + 2 = 3b
To collect the variable terms on one side, subtract 3b from both sides.
–3b –3b
b + 2 = 0
Since 2 is added to b, subtract 2 from both sides.
– 2 – 2
b = –2

8. Check It Out! Example 1a Continued
Solve the equation. Check your answer.
Check
4b + 2 = 3b
To check your answer, substitute –2 for b.
4(–2) (–2)
– –6 
–6 –6

9. Check It Out! Example 1b
Solve the equation. Check your answer.
y = 0.7y – 0.3
To collect the variable terms on one side, subtract 0.3y from both sides.
y = 0.7y – 0.3
–0.3y –0.3y
= 0.4y – 0.3
Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction.
= 0.4y
Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.
2 = y

10. Check It Out! Example 1b Continued
Solve the equation. Check your answer.
Check
y = 0.7y – 0.3
To check your answer, substitute 2 for y.
(2) 0.7(2) – 0.3
– 0.3 

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

12. Additional Example 2: Simplifying Each Side Before Solving Equations
Solve the equation.
4 – 6a + 4a = –1 – 5(7 – 2a)
4 – 6a + 4a = –1 –5(7 – 2a)
Distribute –5 to the expression in parentheses.
4 – 6a + 4a = –1 –5(7) –5(–2a)
4 – 6a + 4a = –1 – a
Combine like terms.
4 – 2a = – a
Since –36 is added to 10a, add 36 to both sides.
40 – 2a = a

13. Additional Example 2 Continued
Solve 4 – 6a + 4a = –1 – 5(7 – 2a).
40 – 2a = 10a
To collect the variable terms on one side, add 2a to both sides.
+ 2a +2a
= 12a
Since a is multiplied by 12, divide both sides by 12.

14. Check It Out! Example 2a
Solve the equation. Check your answer.
Distribute to the expression in parentheses. 1 2
To collect the variable terms on one side, subtract b from both sides. 1 2
3 = b – 1
Since 1 is subtracted from b, add 1 to both sides.
4 = b

15. Check It Out! Example 2a Continued
Solve the equation. Check your answer.
Check
To check your answer, substitute 4 for b. 
5 5

16. Check It Out! Example 2b
Solve the equation. Check your answer.
3x + 15 – 9 = 2(x + 2)
3x + 15 – 9 = 2(x + 2)
Distribute 2 to the expression in parentheses.
3x + 15 – 9 = 2(x) + 2(2)
3x + 15 – 9 = 2x + 4
Combine like terms.
3x + 6 = 2x + 4
To collect the variable terms on one side, subtract 2x from both sides.
–2x –2x
x + 6 =
Since 6 is added to x, subtract 6 from both sides.
– – 6
x = –2

17. Check It Out! Example 2b Continued
Solve the equation. Check your answer.
Check
3x + 15 – 9 = 2(x + 2)
3(–2) + 15 – (–2 + 2)
To check your answer, substitute –2 for x.
– – (0)
0 0 

18. 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.

19. Additional Example 3A: Infinitely Many Solutions or No Solutions
Solve the equation.
10 – 5x + 1 = 7x + 11 – 12x
Identify like terms.
10 – 5x + 1 = 7x + 11 – 12x
Combine like terms on the left and the right.
11 – 5x = 11 – 5x
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.

20. Additional Example 3B: Infinitely Many Solutions or No Solutions
Solve the equation.
12x – 3 + x = 5x – 4 + 8x
12x – 3 + x = 5x – 4 + 8x
Identify like terms.
Combine like terms on the left and the right.
13x – 3 = 13x – 4
–13x –13x
Subtract 13x from both sides. 
–3 = –4
False statement; the solution set is .
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.

21. The empty set can be written as  or {}.
Writing Math

22.  Check It Out! Example 3a Solve the equation. 4y + 7 – y = 10 + 3y
Identify like terms.
3y + 7 = 3y + 10
–3y –3y
Subtract 3y from both sides. 
False statement; the solution set is .
7 = 10
The equation 4y + 7 – y = y is always false. There is no value of y that will make the equation true. There are no solutions.

23. Check It Out! Example 3b
Solve the equation.
2c c = –14 + 3c + 21
2c c = –14 + 3c + 21
Identify like terms.
Combine like terms on the left and the right.
3c + 7 = 3c + 7
The statement 3c + 7 = 3c + 7 is true for all values of c. The equation 2c c = –14 + 3c + 21 is an identity. All values of c will make the equation true. In other words, all real numbers are solutions.

24. Additional Example 4: Application
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?
Person
Bulbs
Jon
60 bulbs plus 44 bulbs per hour
Sara
96 bulbs plus 32 bulbs per hour

25. Additional Example 4 Continued
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
32 bulbs each hour
When is ?
h = h
h = h
To collect the variable terms on one side, subtract 32h from both sides.
–32h –32h
h = 96

26. Additional Example 4 Continued
h = 96
Since 60 is added to 12h, subtract 60 from both sides.
– – 60
12h = 36
Since h is multiplied by 12, divide both sides by 12 to undo the multiplication.
h = 3

27. 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 h = 3:
h = (3) = = 192
h = (3) = = 192
After 3 hours, Jon and Sara will each have planted 192 bulbs.

28. Let g represent Greg's age, and write expressions for his age.
Check It Out! Example 4
Four times Greg's age, decreased by 3 is equal to 3 times Greg's age, increased by 7. How old is Greg?
Let g represent Greg's age, and write expressions for his age.
Four times Greg's age
three times Greg's age
is equal to
decreased by
increased by 3 7 .
4g – = g

29. Check It Out! Example 4 Continued
To collect the variable terms on one side, subtract 3g from both sides.
4g – 3 = 3g + 7
–3g –3g
g – 3 =
Since 3 is subtracted from g, add 3 to both sides.
g =
Greg is 10 years old.

30. Lesson Quiz 1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4
Solve each equation.
1. 7x + 2 = 5x (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