## Section4.1The Distributive Law

### SubsectionProducts

So far we have considered sums and differences of algebraic expressions. Now we take a look at simple products. To simplify a product such as $3(2a)\text{,}$ we multiply the numerical factors to find

\begin{equation*} 3(2a)=(3 \cdot 2)a = 6a \end{equation*}

This calculation is an application of the associative law for multiplication. We can see why the product is reasonable by recalling that multiplication is repeated addition, so that

\begin{equation*} 3(2a) = 2a + 2a + 2a = 6a~~~~~~~~~~~~~\blert{\text{Three terms}} \end{equation*}
###### Example4.1.
1. $5(3x)=15x$
2. $-2(-4h)=8h$

###### 1.

Explain the difference between the expressions $2b+3b$ and $2(3b)\text{.}$

### SubsectionThe Distributive Law

Another useful property for simplifying expressions is the distributive law. First consider a numerical example. Suppose we would like to find the area of the two rooms shown at right. We can do this in two different ways:

• Method 1: Think of the rooms as one large rectangle, with width 10 feet and length $12+8$ feet. The area is then given by

\begin{equation*} \begin{aligned} \text{Area} \amp = \text{width} \times \text{length}\\ \amp = \blert{10(12+8)} = 10(20) = 200~\text{square feet} \end{aligned} \end{equation*}
• Method 2: Find the area of each room separately and add the two areas together. This gives us

\begin{equation*} \blert{10(12) + 10(8)} = 120+80 = 200~\text{square feet} \end{equation*}

Of course, we get the same answer with either method. However, looking at the first step in each calculation, we see that

\begin{equation*} 10(12+8)= 10(12) + 10(8) \end{equation*}

We see that there are two ways to compute the expression $10(12+8)\text{.}$

• The first way follows the order of operations, adding $12+8$ first.
• The second way distributes the multiplication to each term inside parentheses; that is, we multiply each term inside the parentheses by 10.

\begin{equation*} \blert{10}(12+8)= \blert{10}(12) + \blert{10}(8) \end{equation*}

This last equation is an example of the distributive law.

###### Distributive Law.

If $a,~ b,$ and $c$ are any numbers, then

\begin{equation*} \blert{a(b+c) = ab + ac} \end{equation*}

If the terms inside parentheses are not like terms, we have no choice but to use the distributive law to simplify the expression.

###### Example4.2.

Simplify $~-2(3x-1)$

Solution

We multiply each term inside parentheses by $-2\text{:}$

\begin{equation*} \begin{aligned} \blert{-2}(3x-1) \amp = \blert{-2}(3x)\blert{-2}(-1)\\ \amp = -6x+2 \end{aligned} \end{equation*}
###### Caution4.3.

Compare the three similar expressions:

\begin{equation*} -3x-5x,~~~~~~-3(-5x),~~~~~~-3(-5-x) \end{equation*}

To simplify these expressions, we must first recognize the operations involved.

• The first is a sum of terms.

\begin{equation*} -3x-5x = -8x \end{equation*}
• The second is a product.

\begin{equation*} -3(-5x) = 15x \end{equation*}
• The third is an application of the distributive law.

\begin{equation*} -3(-5-x) = 15+3x \end{equation*}

###### 2.

When do we need the distributive law to simplify an expression?

### SubsectionSolving Equations and Inequalities

If an equation or inequality contains parentheses, we use the distributive law to simplify each side before we begin to solve.

###### Example4.4.

Solve $~25-6x=3x-2(4-x)$

Solution

We apply the distributive law to the right side of the equation.

\begin{equation*} \begin{aligned} 25-6x \amp = 3x-2(4-x) \amp\amp \blert{\text{Apply the distributive law.}}\\ 25-6x \amp = \blert{3x} - 8 + \blert{2x} \amp\amp \blert{\text{Combine like terms.}}\\ 25-6x \amp = 5x-8 \amp\amp \blert{\text{Subtract}~5x~ \text{from both sides.}}\\ 25-11x \amp = -8 \amp\amp \blert{\text{Subtract 25 from both sides.}}\\ -11x \amp = -33 \amp\amp \blert{\text{Divide both sides by}~-11.}\\ x \amp =3 \amp\amp \blert{\text{The solution is 3.}} \end{aligned} \end{equation*}

You should verify each step of the solution, and check the answer.

###### Caution4.5.

The expression on the right side of the equation in Example 4.4 is made up of two terms:

\begin{equation*} 3x~~~~~~\text{and}~~~~~~{-2}(4-x) \end{equation*}

Remember that we think of all terms as added together, so the minus sign in front of the 2 tells us that 2 is negative. When we apply the distributive law, we multiply each term inside parentheses by $-2$ to get $-8+2x\text{.}$

We know that a fraction bar often serves as a grouping device, like parentheses.

###### Example4.6.

Solve the proportion $~\dfrac{7.6}{1.2} = \dfrac{x+3}{2x-1}$

Solution

We apply the property of proportions and cross-multiply to get

\begin{equation*} \begin{aligned} 7.6(2x-1) \amp = 1.2(x+3) \amp\amp \blert{\text{Apply the distributive law.}}\\ 15.2x-7.6 \amp = 1.2x+3.6 \amp\amp \blert{\text{Subtract}~1.2x~ \text{from both sides.}}\\ 14x-7.6 \amp = 3.6 \amp\amp \blert{\text{Add}~7.6~ \text{to both sides.}}\\ 14x \amp = 11.2 \amp\amp \blert{\text{Divide both sides by 14.}}\\ x \amp = \dfrac{11.2}{14} = 0.8 \end{aligned} \end{equation*}

We can check the solution by substituting $x=0.8$ into the original proportion.

###### 3.

In Example 4.6, why did we put parentheses around $2x-1$ and $x+3\text{?}$

### SubsectionAlgebraic Expressions

We can often simplify algebraic expressions by combining like terms.

###### Example4.7.

Delbert and Francine are collecting aluminum cans to recycle. They will be paid $x$ dollars for every pound of cans they collect. At the end of three weeks, Delbert collected 23 pounds of aluminum cans, and Francine collected 47 pounds.

1. Write algebraic expressions for the amount of money Delbert made, and the amount Francine made.
2. Write and simplify an expression for the total amount of money Delbert and Francine made from aluminum cans.
Solution
1. We multiply the number of pounds collected by the price per pound. Delbert made $23x$ dollars, and Francine made $47x$ dollars.

\begin{equation*} 23x+47x=70x \end{equation*}

###### 4.

In Example 4.7, what does $x$ stand for? What does $23x$ stand for?

The distributive law is also helpful for writing algebraic expressions.

###### Example4.8.

The length of a rectangle is 3 feet less than twice its width, $w\text{.}$

1. Write an expression for the length of the rectangle in terms of $w\text{.}$
2. Write and simplify an expression for the perimeter of the rectangle in terms of $w\text{.}$
3. Suppose the perimeter of the rectangle is 36 feet. Write and solve an equation to find the dimensions of the rectangle.
Solution
1. "Three feet less than twice the width" tells us to subtract 3 from twice the width. The length of the rectangle is $2w-3\text{.}$
2. The perimeter of a rectangle is given by the formula $P=2l+2w\text{.}$ We substitute our expression for the length to get

\begin{equation*} \begin{aligned} P \amp = 2(\alert{2w-3}) +2w \amp\amp \blert{\text{Apply the distributive law.}}\\ \amp = 4w-6+2w \amp\amp \blert{\text{Combine like terms.}}\\ \amp = 6w-6 \end{aligned} \end{equation*}
3. We set our expression for the perimeter equal to 36.

\begin{equation*} \begin{aligned} 6w-6 \amp = 36 \amp\amp \blert{\text{Add 6 to both sides.}}\\ 6w \amp = 42 \amp\amp \blert{\text{Divide both sides by 6.}}\\ w \amp = 7 \end{aligned} \end{equation*}

The width of the rectangle is $\alert{7}$ feet, and its length is $2(\alert{7})-3=11$ feet.

###### 5.

In Example 4.8, why did we multiply $2w-3$ by 2?

### SubsectionSkills Warm-Up

#### ExercisesExercises

For Problems 1-2, use a formula to write an equation. Then solve the equation.

###### 1.

32.5% of the class are engineering majors. If there are 91 engineering majors, how many students are in the class?

###### 2.

A rectangular cookie sheet is 40 cm long and has a perimeter of 116 cm. How wide is the cookie sheet?

For Problems 3-4, write an equation you could solve to answer the question.

###### 3.

Erika bought a 50-pound bag of dog food and after 28 days she had 29 pounds left. How much dog food did she use per day?

###### 4.

A hamburger contains 60 calories less than two bags of fries. A hamburger and one bag of fries contains 780 calories. How many calories are in a bag of fries?

### ExercisesHomework 4.1

For Problems 1–4, use the distributive law to remove parentheses. For some of these exercises you will use the distributive law in the form

\begin{equation*} \blert{(b+c)a=ba+ca} \end{equation*}
###### 1.
$5(2y-3)$
###### 2.
$-2(4x+8)$
###### 3.
$-(5b-3)$
###### 4.
$(-6+2t))-6)$

For Problems 5–6, simplify each product. Which product in each pair requires the distributive law?

###### 5.
1. $8(4c)$
2. $8(4+c)$
###### 6.
1. $2(-8-t)$
2. $2(-8t)$
###### 7.

Which of the following is a correct application of the distributive law?

1. $5(3a)=15a$
2. $5(3+a)=15+a$
3. $5(3a)=15(5a)$
4. $5(3+a)=15+5a$
###### 8.

Simplify each expression if possible. Then evaluate for $x=3,~y=9\text{.}$

1. $2(xy)$
2. $2(x+y)$
3. $2-xy$
4. $-2xy$

For Problems 9–12, simplify and combine like terms.

###### 9.
$-6(x+1)+2x$
###### 10.
$5-2(4x-9)+9x$
###### 11.
$-4(3+2z)+2z-3(2z+1)$
###### 12.
$3(-3a-4)-3a-4(3a-4)$

For Problems 13–20, solve the equation or inequality.

###### 13.
$6(3y-4)=-60$
###### 14.
$5w-64=-2(3w-1)$
###### 15.
$-22c+5(3c+4) \gt -2(4-3t)$
###### 16.
$6-4(2a+3) \ge 6a + 2$
###### 17.
$4-3(2t-4) \gt -2(4-3t)$
###### 18.
$0.25(x+3)-0.45(x-3)=0.30$
###### 19.
$\dfrac{a}{a+2}=\dfrac{2}{3}$
###### 20.
$\dfrac{0.3}{0.5} = \dfrac{b+2}{12-b}$
###### 21.

Choose a value for the variable and show that the following pairs of expressions are not equivalent.

1. $5(x+3)$ and $5x+3$
2. $-4(c-3)$ and $-4c-12$
###### 22.
1. Evaluate the expression $(2x+7)-4(4x-2)-(-2x+3)$ for $x=-3\text{.}$
2. Simplify the expression in part (a) by combining like terms.
3. Evaluate your answer to part (b) for $x=-3\text{.}$ Check that you got the same answer for part (a).
4. Are the expressions in parts (a) and (b) equivalent? Why or why not?
###### 23.

The length of a rectangle is 8 feet shorter than twice its width. Write expressions in terms of $w\text{,}$ the width of the rectangle.

1. The length of the rectangle
2. The perimeter of the rectangle
3. The area of the rectangle

###### 33.

An apple and a glass of milk together contain 260 calories.

1. If an apple contains $a$ calories, how many calories are in a glass of milk?
2. Write an expression for the number of calories in two apples and three glasses of milk.
3. If two apples and three glasses of milk contain 660 calories, find the number of calories in an apple.
###### 34.

Melody sold 47 tickets to a charity concert. Reserved seats cost $10 and open seating was$6 a ticket. Let $x$ represent the number of reserved seats she sold. Write expressions in terms of $x$ for:

1. The number of open seating tickets Melody sold
2. The amount of money Melody collected from reserved seating tickets
3. The amount of money Melody collected from open seating tickets
4. The total amount of money Melody collected
###### 35.

Melody from Problem 34 collected \$330 from the sale of concert tickets. Write and solve an equation to answer the question: How many reserved seats did she sell?