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Section 4.1 The Distributive Law

Subsection Products

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*}
Example 4.1.

  1. \(\displaystyle 5(3x)=15x\)

  2. \(\displaystyle -2(-4h)=8h\)

Reading Questions Reading Questions

1.

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

Answer.

The first expression is a sum and the second one is a product.

Subsection The 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:

rectangle

  • 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{align*} \text{Area} \amp = \text{width} \times \text{length}\\ \amp = \blert{10(12+8)} = 10(20) = 200~\text{square feet} \end{align*}

  • 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*}
Look Ahead.

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

Example 4.2.

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

Solution.

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

\begin{align*} \blert{-2}(3x-1) \amp = \blert{-2}(3x)\blert{-2}(-1)\\ \amp = -6x+2 \end{align*}
Caution 4.3.

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*}

Reading Questions Reading Questions

2.

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

Answer.

If the terms inside parentheses are not like terms

Subsection Solving Equations and Inequalities

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

Example 4.4.

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

Solution.

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

\begin{align*} 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{align*}

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

Caution 4.5.

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.

Example 4.6.

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{align*} 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{align*}

We can check the solution by substituting \(x=0.8\) into the original proportion.

Reading Questions Reading Questions

3.

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

Answer.

To replace the fraction bars as grouping symbols

Subsection Algebraic Expressions

We can often simplify algebraic expressions by combining like terms.

Example 4.7.

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.

  2. We add the amount of money Delbert made to the amount Francine made.

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

Reading Questions Reading Questions

4.

In Example 4.7, what does \(x\) stand for? What does \(23x\) stand for?

Answer.

The fee for a pound of recycled cans; the amount of money Delbert made

The distributive law is also helpful for writing algebraic expressions.

Example 4.8.

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{align*} 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{align*}

    >

  3. We set our expression for the perimeter equal to 36.

    \begin{align*} 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{align*}

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

Reading Questions Reading Questions

5.

In Example 4.8, why did we multiply \(2w-3\) by 2?

Answer.

We substituted into the formula for perimeter.

Subsection Skills Warm-Up

Exercises Exercises

Exercise Group.

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?

Exercise Group.

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?

Subsubsection Answers to Skills Warm-Up

  1. \(\displaystyle 0.325x=91;~~280\)

  2. \(2w+2(40)=116;~~18~\)cm

  3. \(50-28d=29,~~\dfrac{3}{4}~\)lb

  4. \(\displaystyle x+(2x-60)=780;~~280\)

Subsection Lesson

Subsubsection Activity 1: Simplifying Expressions

Exercises Exercises
1.

Simplify each product.

  1. \(\displaystyle 6(-2b)\)

  2. \(\displaystyle -4(-7w)\)

2.

Use the distributive law to simplify each expression.

  1. \(\displaystyle 8(3y-6)\)

  2. \(\displaystyle -3(7+5x)\)

Subsubsection Activity 2: Solving Equations

Steps for Solving Linear Equations.

  1. Use the distributive law to remove any parentheses.

  2. Combine like terms on each side of the equation.

  3. By adding or subtracting the same quantity on both sides of the equation, get all the variable terms on one side and all the constant terms on the other.

  4. Divide both sides by the coefficient of the variable to obtain an equation of the form \(x=a\text{.}\)

Exercises Exercises
1.

Solve \(~~6-3(x-2)=4x-(x+8)\)

2.

Solve \(~~2n+(4-3n) \ge 6-(3-2n)\)

Subsubsection Activity 3: Problem Solving

Exercises Exercises
1.

Shalia runs a landscaping business. She has a budget of $385 to buy 20 rose bushes for one of her clients. Hybrid tea roses cost $21 each, and shrub roses cost $16.

  1. If Shalia buys tea roses, write an expression for the number of shrub roses she needs.

  2. Write expressions for the cost of the tea roses and the cost of the shrub roses.

  3. Write and simplify an expression for the total cost of the roses.

2.

One angle of a triangle is three times the smallest angle, and the third angle is 20 greater than the smallest angle. Find the degree measure of each angle.

  • Step 1 Let \(x\) represent the smallest angle, and write expressions for the other two angles.

    \begin{gather*} \blert{\text{Second angle:}}\\ \blert{\text{Third angle:}} \end{gather*}

  • Step 2 Write an equation, using the fact that the sum of the three angles of a triangle is 180\(\degree\text{.}\)

  • Step 3 Solve the equation. Begin by simplifying the left side.

  • Write your answer in a sentence.

Subsubsection Wrap-Up

Objectives.

In this Lesson we practiced the following skills:

  • Applying the distributive law

  • Solving linear equations

  • Writing equations for applied problems

Questions.

  1. Explain the difference between \(6(-2b)\) and \(6(-2+b)\text{.}\) Which law or property do you apply to simplify each expression?

  2. Explain how to simplify each side of an equation before beginning to solve.

  3. What formula did you use in Problem 2 of Activity 3?

Subsection Homework Preview

Exercises Exercises

Exercise Group.

Simplify.

1.
\(8-3(2x+4)-3x-2\)
2.
\(3a-4-3(2a-5)\)
3.

  1. \(\displaystyle -4(-6t)+2\)

  2. \(\displaystyle -4(-6t+2)\)

  3. \(\displaystyle -4-6t-2\)

  4. \(\displaystyle (-4-6t)(-2)\)

Exercise Group.

Solve.

4.
\(-9+2(9+4z)=-23\)
5.
\(35=4(2w+5)-3w\)
6.
\(4(2-3w)=9-3(2w-1)\)

Subsubsection Answers to Homework Preview

  1. \(\displaystyle -9x-6\)

  2. \(\displaystyle -3a+11\)

    1. \(\displaystyle 24t+2\)

    2. \(\displaystyle 24t-8\)

    3. \(\displaystyle -6-6t\)

    4. \(\displaystyle 8+12t\)

  4. \(\displaystyle -4\)

  5. \(\displaystyle 3\)

  6. \(\displaystyle \dfrac{-2}{3}\)

Exercises Homework 4.1

Exercise Group.

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.
\((-6+2t))-6)\)
Exercise Group.

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

5.

  1. \(\displaystyle 8(4c)\)

  2. \(\displaystyle 8(4+c)\)

6.

  1. \(\displaystyle 2(-8-t)\)

  2. \(\displaystyle 2(-8t)\)

7.

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

  1. \(\displaystyle 5(3a)=15a\)

  2. \(\displaystyle 5(3+a)=15+a\)

  3. \(\displaystyle 5(3a)=15(5a)\)

  4. \(\displaystyle 5(3+a)=15+5a\)

8.

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

  1. \(\displaystyle 2(xy)\)

  2. \(\displaystyle 2(x+y)\)

  3. \(\displaystyle 2-xy\)

  4. \(\displaystyle -2xy\)

Exercise Group.

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.
\(3(-3a-4)-3a-4(3a-4)\)
Exercise Group.

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

24.

Risa bought a skirt and a sweater that together cost $108. Then she bought a pair of shoes that cost twice as much as the sweater. Write and simplify expressions in terms of \(c\text{,}\) the cost of the skirt:

  1. The cost of the sweater

  2. The cost of the shoes

  3. The total amount Risa spent

25.

Albert and Isaac left the same hotel at the same time, and each drove for 3 hours, but Albert drove 15 miles per hour faster than Isaac. Write and simplify expressions in terms of Isaac's speed, \(s\text{.}\)

  1. How far Isaac drove

  2. How far Albert drove

  3. If they drove in the same direction, how far apart they are now

  4. If they drove in opposite directions, how far apart they are now

26.

A box of AlmondOats contains 15 ounces of cereal made of oat flakes and sliced almonds. Oat flakes cost 15 cents per ounce, and almonds cost 35 cents per ounce. Write and simplify expressions in terms of \(a\text{,}\) the number of ounces of almonds in the box.

  1. The number of ounces of oat flakes in the box

  2. The cost of the almonds

  3. The cost of the oat flakes

  4. The total cost of a box of AlmondOats

27.

Trader Jim's Pomegranate Punch is 25% pomegranate juice, and Pomajoy is 60% pomegranate juice. Vera mixes some of each to make 8 quarts of juice drink. Write and simplify expressions in terms of \(x\text{,}\) the number of quarts of Pomajoy she uses.

  1. The number of quarts of Pomegranate Punch

  2. The number of quarts of pomegranate juice in the Pomajoy

  3. The number of quarts of pomegranate juice in the Pomegranate Punch

  4. The number of quarts of pomegranate juice in Vera's mixture

28.

Premium ice cream is 12% butterfat, and chocolate syrup is 55% butterfat. A chocolate sundae, without whipped cream and a cherry, weighs 10 ounces. Write and simplify expressions in terms of \(x\text{,}\) the number of ounces of ice cream in the sundae.

  1. The number of ounces of chocolate syrup in the sundae

  2. The amount of fat in the ice cream

  3. The amount of fat in the chocolate syrup

  4. The total amount of fat in the sundae

Exercise Group.

For Problems 29–30, find an algebraic expression for the area of the rectangle. Use the distributive law to write the expression in two ways.

29.
rectangle
30.
rectangle
31.

The length of a rectangular vegetable garden is 6 yards more than twice its width.

  1. Write an expression for the perimeter of the garden in terms of its width.

  2. Ann bought 42 yards of fence to enclose the garden. What are its dimensions?

32.

Revenue from the state lottery is divided between education and administrative costs in the ratio of 4 to 3. If the lottery revenue this year is $24,000,000, how much money will go to education?

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?