Chapter 2: Linear Equations and Ineqalities
2.1A: Using the Distributive Property
Goals for Class today:
 Analyze and solve verbal problems whose solution requires solving a linear
equation in one variable or linear inequality in one variable
 Solve all types of linear equations in one variable
Big Idea:
When using the distributive property, it is very important to remember to multiply every
inside term by the outside term.
outside term  inside term  inside term
inside term 
inside term outside term
The outside term can come before or after the inside terms.
3 (a+b) = 3a + 3b
3 (a-b) = 3a - 3b
(a + b)3 = 3a + 3b
(a - b)3 = 3a - 3b
7( a + b + c - d + e) = 7a + 7b +7c - 7d + 7e
What is the value of n in the equation 0.6(n +10) = 3.6?
(1) –0.4
(3) –4
(2) 5
(4) 4
One Solution
Use the distributive property’
Given:
0.6(n+10) = 3.6
Check
DistProp
.6(n)+.6(10)=3.6
.6(-4)+.6(10)=3.6
.6n+ 6 =3.6
-2.4 + 6 =3.6
S 6
-6
-6
3.6
=3.6
D  .6 
Answer
.6n
.6n
.6
n
=-2.4
-2.4
=
.6
= -4
The answer is (3) -4
1
T- 1 -
Chapter 2: Linear Equations and Ineqalities
2.1A: Using the Distributive Property
A candy store sells 8-pound bags of mixed hazelnuts and cashews. If c pounds of
cashews are in a bag, the price p of the bag can be found using the formula
p  2.59c  1.72(8  c). If one bag is priced at $18.11, how many pounds of cashews
does it contain?
One Solution
A formula is provided and a price is provided. What the problem is asking us to do is
solve for c when the price is $18.11. Once we understand this, the rest is easy.
Given:
p=2.59c+1.72(8-c)
Given:
Therefore:
p=18.11
18.11=2.59c+1.72(8-c)
Dist.Prop.
18.11=2.59c+1.72(8)-1.72(c)
18.11=2.59c+13.76-1.72c
Rewrite:
CLT:
18.11=2.59c-1.72c+13.76
18.11=(2.59-1.72)c+13.76
18.11=.87c+13.76
Subtract 13.76
Divide.87c 
Answer
8.11-13.76=.87c
4.35=.87c
4.35 .87c
=
.87 .87
5=c
2
Check
18.11=2.59  5  +1.72(8-5)
18.11=12.95+1.72  3
18.11=12.95+5.16
18.11=18.11
T- 2 -
Chapter 2: Linear Equations and Ineqalities
T- 3 -
2.1A: Using the Distributive Property
1
Chad complained to his friend that he Not all of the homework problems are
had five equations to solve for equations. The first problem is an
homework. Are all of the homework expression.
problems equations?
Justify your
answer.
2
The expression 2 x 2  x 2 is equivalent to
(1) x 0
(3) x 2
(2) 2
(4)  2x 4
3
The expression
3x
5
5x 2
(3)
10
(1)
5x x
 is equivalent to
6 4
13x
(2)
12
5x
(4)
24
3
(3)
(2)
(5 x  4)  (6  x)
(6)( 4)
20 x  6 x
24
26 x
24
13 x
12
Chapter 2: Linear Equations and Ineqalities
T- 4 -
2.1A: Using the Distributive Property
4
d
and
2
expressed in simplest form?
3d
7d
(1)
(3)
5
5
3d
7d
(3)
(4)
6
6
What is the sum of
2d
3
4
(4)
(d  3)  (2  2d )
(2)(3)
3d  4d
6
7d
6