5 x 4 5x + 20 5x 20

Name: __________________________
Period: _______
Week 27 Homework Packet – 7th Grade Math Honors
Score
/4
ASSIGNMENT DUE TUESDAY, FEBRUARY 28, 2017
Unit 4, Chapter 7 – Algebraic Expressions
Use the distributive property to simplify each
expression below.
1) 4(π‘₯ + 5)
Week Score
/12
Stamp
For each problem below, use boxes to describe
how the distributive property can be used to
simplify an expression, then write the simplified
expression. An example has been done for you.
Ex.) 5(π‘₯ + 4)
x
4
5x
20
2) βˆ’3(π‘š + 2)
5
3) π‘₯(𝑦 + 3)
5x + 20
11) 3(π‘š + 7)
4) βˆ’5(π‘ž βˆ’ 4)
5) 2(π‘₯ + 𝑦 + 4)
12) 4(π‘˜ βˆ’ 2)
6) 3(2π‘₯ βˆ’ 3𝑦)
7) βˆ’4(2π‘š βˆ’ 5𝑛)
8)
1
2
13) 2(π‘₯ + 𝑦 + 5)
(4π‘₯ βˆ’ 6𝑦 + 3)
9) 2π‘Ž(4𝑏 + 3𝑐 βˆ’ 5)
14) 3(π‘Ž + 𝑏 βˆ’ 1)
10) βˆ’(2𝑐 βˆ’ 4𝑑)
Page 1
Corrected by: _______________________________
Name: __________________________
7th Grade Math Honors
Period: _______
Score
/4
ASSIGNMENT DUE WED/THU, MARCH 1/2, 2017
Unit 4, Chapter 7 – Algebraic Expressions
Simplify each expression below by combining like
terms. Show all steps.
1) 3π‘Ž + 7π‘Ž + π‘Ž + 9 + 10
Stamp
Skills Review
Add or subtract the following fractions by finding
the least common denominator. Show all work.
11)
2
12)
1
13)
3
5
+
1
4
2) 𝑏 + 7 + 4𝑏 + 10
3) 3π‘š + 8𝑛 + 2π‘š βˆ’ 3𝑛
3
+
1
6
4) 3𝑐 + 7𝑑 + 2 + 5𝑐 + 10𝑑
5) 9β„Ž βˆ’ 3β„Ž + 10π‘˜ βˆ’ π‘˜ + β„Ž
7
βˆ’
1
2
6) 3π‘Ž + 2(π‘Ž + 7)
14)
2
3
βˆ’
1
8
7) 4(π‘₯ + 7) βˆ’ 4π‘₯
8) βˆ’2(π‘š βˆ’ 4) + 3π‘š
15)
3
1
+ (βˆ’ )
4
8
9) 5 βˆ’ 2(π‘₯ + 4)
10) 10𝑔 βˆ’ 3(βˆ’2𝑔 βˆ’ 6)
Page 2
16)
5
6
βˆ’
3
4
Corrected by: _______________________________
Name: _________________________________
7th Grade Math Honors
Period: _________
Score
ASSIGNMENT DUE FRIDAY, MARCH 3, 2017
Unit 4, Chapter 7 – Algebraic Expressions
Add the following expressions together and write
your answer in simplified form.
Stamp
/4
Simplify each of the following using all of the
properties of algebra that we have used so far.
Show each step.
1) (2π‘₯ + 4) + (3π‘₯ + 7)
9) 4(8π‘š + 6) + (π‘š βˆ’ 2)
2) (4π‘š βˆ’ 3) + (5π‘š + 7)π‘š
10) βˆ’3(7 βˆ’ 2π‘₯) + 7(π‘₯ βˆ’ 2)
3) (βˆ’3π‘Ž + 9) + (10π‘Ž βˆ’ 7)
4) (4π‘Ÿ βˆ’ 3) + (π‘Ÿ + 3)
11) 5(3π‘Ž + 1) + 3(2 βˆ’ 4π‘Ž) βˆ’ 3π‘Ž
5) (2π‘˜ + 11) + (βˆ’8π‘˜ βˆ’ 4)
6) (7𝑀 + 3) + (8𝑀 βˆ’ 19)
12) 3(π‘Ÿ + 7) + 7(2π‘Ÿ βˆ’ 3)
7) (2π‘Ž βˆ’ 4𝑏 + 7) + (3π‘Ž βˆ’ 10)
13) 3(4 βˆ’ 9𝑛) βˆ’ 2(5𝑛 βˆ’ 3)
8) (3π‘₯ βˆ’ 7𝑦 + 10) + (βˆ’2π‘₯ + 6𝑦 βˆ’ 8)
Page 3
Corrected by: _______________________________