PRE-ALGEBRA

PRE-ALGEBRA

Lesson 7-2 Warm-Up

PRE-ALGEBRA

“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: 0.5a2 + 0.875 = 13.25

Since 0.875 is the greatest number of place values from the end (3), jump all of the decimals 3 places to the right, so 0.5a2 + 0.875 = 13.25 is the same as 500a2 + 875 = 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 • 20 + 4 • 5 = 80 + 20 = 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 – 7 + 7 = 20 + 7

Step 5: Undo multiplication and division.

Examples: 2x = 10 • = 20 4

2 2

• • •

t4

t4

41 •

PRE-ALGEBRA

Example: Solve 2c + 2 + 3c = 12.

2c + 2 + 3c = 12

2c + 2 + 3c = 12Check:

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 5

= Isolate the variable. Use the Division Property of Equality.

c = 2 Simplify.

2(2) + 2 + 3(2) 12 Substitute 2 for c.

12 = 12 The solution checks.

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

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 = 41 Combine like terms.

4s + 5 – 5 = 41 – 5   Subtract 5 from each side.

4s = 36 Simplify.

s = 9 Simplify.

= Divide each side by 4.4s4

364

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

(continued)

Check:   Is the solution reasonable? Helen and Jorge have a total of 41 stamps. Since 9 + 32 = 41, the solution is reasonable.

Helen has 9 stamps. Jorge has 3(9) + 5 = 32 stamps.

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

Example: Solve 2(3a + 6) + a = 110

7a + 12 = 110 Combine like terms (3 + 1).

-12 - 12 Subtract 12 from each side.

7a = 98 Simplify.

2(3a + 6) + a = 110Check:

110 = 110

2(48) + 14 90

2(3 • 14 + 6) + 14 110 Substitute 14 for a.

a = 14 Simplify.

= Divide each side by 7. 7a7

987

Solving Multi-Step Equations

Additional Examples

6a + 1a + 12 = 110 Commutative Property of Addition (Note: a = 1a Identity Property)

LESSON 7-2

6a + 12 + 1a = 110 Distributive Property 2(3a + 6) = 2 • 3a + 2 • 6

1

1 1

14

PRE-ALGEBRA

Solve each equation.

a. 4(2q – 7) = –4

4(2q – 7) = –4

8q – 28 = –4 Use the Distributive Property.

8q – 28 + 28 = –4 + 28 Add 28 to each side.

8q = 24 Simplify.

q = 3 Simplify.

Divide each side by 8.=8q8

248

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

(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 – 20 = –6r + 20 – 20 Subtract 20 from each side.

24 = –6r Simplify.

–4 = r Simplify.

Divide each side by –6.=24–6

–6r–6

44 = –6r + 20 Combine like terms (r = 1r by the Identity Property).

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

Example: The sum of three consecutive integers is 96. Find

the integers.

sum of three consecutive integers 96isWords

Let = the least integer.n

Then = the second integer,n + 1

and = the third integer.n + 2

+ +n n + 1 n + 2Equation 96=

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

(continued)

n + (n + 1) + (n + 2) = 96 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.

n = 31 Simplify.

= Divide each side by 3.3n3

963

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

(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 31 + 32 + 33 = 96.

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

The sum of three consecutive integers is 42.

Find the integers.

sum of three consecutive integers 42isWords

Let = the least integer.n

Then = the second integer,n + 1

and = the third integer.n + 2

+ +n n + 1 n + 2Equation 42=

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

(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.

n = 13 Simplify.

= Divide each side by 3.3n3

393

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

(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 13 + 14 + 15 = 42.

Solving Multi-Step EquationsLESSON 7-2

Additional Examples

PRE-ALGEBRA

Solve each equation.

1. b + 2b – 11 = 88 2. 6(2n – 5) = –90 3. 3(x + 6) + x = 86

4. Find four consecutive integers whose sum is –38.

33 –5 17

–11, –10, –9, –8

Solving Multi-Step EquationsLESSON 7-2

Lesson Quiz