2. Lesson 7-2 Warm-Up

3. “Solving Multi-Step Equations” (7-2)
What are the steps for solving a multi-step equation?
Step 1: Clear the equation of fractions and decimals. You can clear decimals by using the decimal that has the most digits after it and moving all of the decimals that number of jumps to the right. Example: 0.5a = Since is the greatest number of place values from the end (3), jump all of the decimals 3 places to the right, so 0.5a = is the same as 500a = 13,250 Step 2: Use the Distributive Property to remove parenthesis if you can’t simplify the problem within them. Example: 4 (25) = 4 (20 + 5) = 4 • • 5 = = 100 Step 3: Combine like terms on each side. Examples: 4x + 5x = 9x 8a – 5a = 3a Step 4: “Undo” (since you’re working backwards) addition and subtraction. Examples: 2x + 5 – 5 = 10 – 5 – = Step 5: Undo multiplication and division. Examples: 2x = 10 • = • • •
t 4
4 1
t 4 •

4. Subtraction Property of Equality
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
Example: Solve 2c c = 12.
2c c = 12
2c + 3c + 2 = 12
Commutative Property
5c + 2 = 12
Combine like terms
5c + 2 – 2 = 12 – 2
Subtraction Property of Equality
5c = 10
Simplify. 5c 5 10 =
Isolate the variable. Use the Division Property of Equality.
c = 2
Simplify.
2c c = 12
Check:
2(2) (2) 12
Substitute 2 for c.
12 = 12
The solution checks.

5. 4s + 5 – 5 = 41 – 5 Subtract 5 from each side.
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
In his stamp collection, Jorge has five more than three times as many stamps as Helen. Together they have 41 stamps. Solve the equation s + 3s + 5 = 41. Find the number of stamps each one has.
s + 3s + 5 = 41
4s + 5 = Combine like terms.
4s + 5 – 5 = 41 – 5   Subtract 5 from each side.
4s = Simplify.
= Divide each side by 4. 4s 4 36
s = Simplify.

6. Helen has 9 stamps. Jorge has 3(9) + 5 = 32 stamps.
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
(continued)
Helen has 9 stamps. Jorge has 3(9) + 5 = 32 stamps.
Check:   Is the solution reasonable? Helen and Jorge have a total of 41 stamps. Since = 41, the solution is reasonable.

7. Example: Solve 2(3a + 6) + a = 110
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
Example: Solve 2(3a + 6) + a = 110
6a a = Distributive Property  (3a + 6) = 2 • 3a + 2 • 6
6a + 1a + 12 = Commutative Property of Addition (Note: a = 1a  Identity Property)
7a + 12 = Combine like terms (3 + 1).
Subtract 12 from each side.
7a = 98 Simplify. 1 14
= Divide each side by 7.
7a 7
98 7 1 1
a = 14 Simplify.
2(3a + 6) + a = 110
Check:
2(3 • ) Substitute 14 for a.
2(48)
110 = 110

8. Use the Distributive Property.
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
Solve each equation.
a. 4(2q – 7) = –4
4(2q – 7) = –4
8q – 28 = –4
Use the Distributive Property.
8q – = –4 + 28
Add 28 to each side.
8q = 24
Simplify.
Divide each side by 8. = 8q 8 24
q = 3
Simplify.

9. Use the Distributive Property.
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
(continued)
b.    44 = –5(r – 4) – r
44 = –5(r – 4) – r
44 = –5r + 20 – r
Use the Distributive Property.
44 = –5r – 1r + 20
Use the Commutative and Associative Properties of Addition to group like terms.
44 = –6r + 20
Combine like terms (r = 1r by the Identity Property).
44 – 20 = –6r + 20 – 20
Subtract 20 from each side.
24 = –6r
Simplify.
Divide each side by –6. = 24 –6
–6r
–4 = r
Simplify.

10. sum of three consecutive integers 96 is Words
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
Example: The sum of three consecutive integers is 96. Find the integers.
sum of three consecutive integers 96 is
Words
Let = the least integer. n
Then = the second integer,
n + 1
and = the third integer.
n + 2 n
n + 1
n + 2
Equation 96 =

11. n + (n + 1) + (n + 2) = 96 Equation
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
(continued)
n + (n + 1) + (n + 2) = Equation
(n + n + n) + (1 + 2) = 96 Use the Commutative and Associative Properties of Addition to group like terms.
3n + 3 = 96 Combine like terms (n = 1n; 1n + 1n + 1n = 3n).
3n + 3 – 3 = 96 – 3   Subtract 3 from each side.
3n = 93 Simplify.
= Divide each side by 3. 3n 3 96
n = 31 Simplify.

12. Check: Is the solution reasonable? Yes, because 31 + 32 + 33 = 96.
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
(continued)
If n = 31, then n + 1 = 32, and n + 2 = 33. The three integers are 13, 14, and 15.
Check: Is the solution reasonable? Yes, because = 96.

13. The sum of three consecutive integers is 42. Find the integers.
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
The sum of three consecutive integers is 42. Find the integers.
sum of three consecutive integers 42 is
Words
Let = the least integer. n
Then = the second integer,
n + 1
and = the third integer.
n + 2 n
n + 1
n + 2
Equation 42 =

14. 3n + 3 – 3 = 42 – 3 Subtract 3 from each side.
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
(continued)
n + (n + 1) + (n + 2) = 42
(n + n + n) + (1 + 2) = 42 Use the Commutative and Associative Properties of Addition to group like terms.
3n + 3 = 42 Combine like terms.
3n + 3 – 3 = 42 – 3   Subtract 3 from each side.
3n = 39 Simplify.
= Divide each side by 3. 3n 3 39
n = 13 Simplify.

15. Check: Is the solution reasonable? Yes, because 13 + 14 + 15 = 42.
Solving Multi-Step Equations
LESSON 7-2
Additional Examples
(continued)
If n = 13, then n + 1 = 14, and n + 2 = 15. The three integers are 13, 14, and 15.
Check: Is the solution reasonable? Yes, because = 42.

16. 4. Find four consecutive integers whose sum is –38. 33 –5 17
Solving Multi-Step Equations
LESSON 7-2
Lesson Quiz
Solve each equation.
1. b + 2b – 11 = (2n – 5) = – (x + 6) + x = 86
4. Find four consecutive integers whose sum is –38. 33 –5 17
–11, –10, –9, –8