7.3 Distributive Property

Learning objectives

Learning objectives

• Simplify expressions using the distributive property

• Evaluate expressions using the distributive property

Simplify Expressions Using the Distributive Property

Suppose three friends are going to the movies. They each need $\text{\textdollar 9.25};$ that is, $9$ dollars and $1$ quarter. How much money do they need all together? You can think about the dollars separately from the quarters.

They need $3$ times $\text{\textdollar 9},$ so $\text{\textdollar 27},$ and $3$ times $1$ quarter, so $75$ cents. In total, they need $\text{\textdollar 27.75}.$

If you think about doing the math in this way, you are using the Distributive Property.

Back to our friends at the movies, we could show the math steps we take to find the total amount of money they need like this:

In algebra, we use the Distributive Property to remove parentheses as we simplify expressions. For example, if we are asked to simplify the expression $3(x+4),$ the order of operations says to work in the parentheses first. But we cannot add $x$ and $4,$ since they are not like terms. So we use the Distributive Property, as shown in Example 7.17.

Example 7.17

Solution

1. Identify the outside factor and the terms. In $3(x+4)$, the factor $3$ multiplies each of the two terms inside the parentheses, $x$ and $4$.
2. Apply the distributive property. Write $3(x+4)$ as $3\cdot x+3\cdot4$, distributing $3$ to both terms.
3. Multiply each product. The products are $3x$ and $12$, so the expanded expression is $3x+12$.

Takeaway: Distributing $3$ across both terms of $x+4$ gives $3x+12$, since each term of the sum is multiplied by $3$.

Some students find it helpful to draw in arrows to remind them how to use the Distributive Property. Then the first step in Example 7.17 would look like this:

Example 7.18

Solution

1. Distribute the factor. Multiply $6$ by each term inside the parentheses: $6(5y+1)=6(5y)+6(1)$.
2. Multiply each product explicitly. The first product is $6\cdot{}5y=30y$, and the second product is $6\cdot{}1=6$.
3. Combine the distributed terms. Writing both products before adding them gives $6(5y)+6(1)=30y+6$. The arrows in the figures show $6$ being sent to both $5y$ and $1$, and the final figure displays the simplified expression.
4. State the result. The simplified expression is $30y+6$.

Takeaway: Always distribute the outside factor to every term inside the parentheses, and write each resulting product before combining them.

The distributive property can be used to simplify expressions that look slightly different from $a(b+c).$ Here are two other forms.

Example 7.19

Solution

1. Distribute the factor over the difference. Rewrite $2(x-3)$ as $2(x)+2(-3)$, so each term inside the parentheses is multiplied by $2$.
2. Multiply each product explicitly. The products are $2\cdot{}x=2x$ and $2\cdot{}(-3)=-6$, which gives $2x-6$.
3. Interpret the figures. The arrows show $2$ multiplying both $x$ and $-3$, the next figure shows the separate products $2x-2\cdot{}3$, and the last figure displays $2x-6$.
4. State the result. The simplified expression is $2x-6$.

Takeaway: When distributing over a subtraction, keep the subtraction sign and multiply the second term by a negative number.

Do you remember how to multiply a fraction by a whole number? We’ll need to do that in the next two examples.

Example 7.20

Solution

1. Distribute the fractional factor. Multiply $\frac{3}{4}$ by each term inside the parentheses: $\frac{3}{4}(n+12)=\frac{3}{4}(n)+\frac{3}{4}(12)$.
2. Multiply each product explicitly. The first product is $\frac{3}{4}\cdot{}n=\frac{3}{4}n$, and the second product is $\frac{3}{4}\cdot{}12=\frac{36}{4}=9$.
3. Combine the two terms. Writing both products before adding gives $\frac{3}{4}n+9$. The figures illustrate the same distribution with arrows and show the product $\frac{3}{4}n$ before the final addition.
4. State the result. The simplified expression is $\frac{3}{4}n+9$.

Takeaway: A fractional outside factor multiplies every term inside the parentheses, so show both products before adding them.

Example 7.21

Solution

1. Distribute the factor. Multiply $8$ by each term inside the parentheses: $8(\frac{3}{8}x+\frac{1}{4})=8(\frac{3}{8}x)+8(\frac{1}{4})$.
2. Multiply each product explicitly. The first product is $8\cdot{}\frac{3}{8}x=3x$, and the second product is $8\cdot{}\frac{1}{4}=2$.
3. Combine the distributed terms. Writing both products before adding them gives $3x+2$. The arrows in the first figure show the factor $8$ reaching both terms, while the later figures display the separate products and the final expression.
4. State the result. The simplified expression is $3x+2$.

Takeaway: Simplify each distributed product on its own before combining the two terms, especially when the factor cancels part of a coefficient.

Using the Distributive Property as shown in the next example will be very useful when we solve money applications later.

Example 7.22

Solution

1. Distribute the factor. Multiply $100$ by each term inside the parentheses: $100(0.3+0.25q)=100(0.3)+100(0.25q)$.
2. Multiply each product explicitly. The first product is $100\cdot{}0.3=30$, and the second product is $100\cdot{}0.25q=25q$.
3. Combine the distributed terms. Writing both products before adding them gives $30+25q$. The figures show arrows from $100$ to both terms, the two separate products, and the simplified expression $30+25q$.
4. State the result. The simplified expression is $30+25q$.

Takeaway: Distributing a whole-number factor over decimals or variables is the same process as distributing it over integers.

In the next example we’ll multiply by a variable. We’ll need to do this in a later chapter.

Example 7.23

Solution

1. Distribute the factor. Multiply $m$ by each term inside the parentheses: $m(n-4)=m(n)+m(-4)$.
2. Multiply each product explicitly. The products are $m\cdot{}n=mn$ and $m\cdot{}(-4)=-4m$, so the expression becomes $mn-4m$.
3. Interpret the figures. The arrows show $m$ multiplying both $n$ and $-4$, the next figure displays the separate products, and the final figure shows the reordered terms $mn-4m$.
4. State the result. The simplified expression is $mn-4m$.

Takeaway: A numerical coefficient is conventionally written before a variable, so $m\cdot{}4$ is written as $4m$.

The next example will use the ‘backwards’ form of the Distributive Property, $(b+c)a=ba+ca.$

Example 7.24

Solution

1. Distribute the outside factor. The factor $p$ multiplies each term inside the parentheses: $(x+8)p=p(x)+p(8)$.
2. Multiply each product explicitly. The products are $p\cdot{}x=px$ and $p\cdot{}8=8p$, so the expression becomes $px+8p$.
3. Interpret the figures. The arrows show $p$ multiplying both $x$ and $8$, and the final figure displays the simplified expression $px+8p$.
4. State the result. The simplified expression is $px+8p$.

Takeaway: The Distributive Property also works in the form $(b+c)a=ba+ca$, so the outside factor may appear on either side.

When you distribute a negative number, you need to be extra careful to get the signs correct.

Example 7.25

Solution

1. Distribute the negative factor. Multiply $-2$ by each term inside the parentheses: $-2(4y+1)=(-2)(4y)+(-2)(1)$.
2. Multiply each product explicitly. The first product is $(-2)\cdot{}4y=-8y$, and the second product is $(-2)\cdot{}1=-2$.
3. Combine the distributed terms. Writing both products before adding them gives $-8y-2$. The figures show the arrows distributing $-2$, then the separate products, and finally the expression $-8y-2$.
4. State the result. The simplified expression is $-8y-2$.

Takeaway: When a negative number is distributed, both resulting products carry its sign, so verify each product separately.

Example 7.26

Solution

1. Distribute the negative factor. Multiply $-11$ by each term inside the parentheses: $-11(4-3a)=(-11)(4)+(-11)(-3a)$.
2. Multiply each product explicitly. The first product is $(-11)\cdot{}4=-44$, and the second product is $(-11)\cdot{}(-3a)=33a$.
3. Combine the distributed terms. Writing both products before adding them gives $-44+33a$. The figures show the distribution, the products $-44-(-11)\cdot{}3a$, and the intermediate form $-44+33a$.
4. State the result. The simplified expression is $-44+33a$, which may also be written as $33a-44$.

Takeaway: Two negative factors produce a positive product, and equivalent simplified expressions may differ only in the order of their terms.

In the next example, we will show how to use the Distributive Property to find the opposite of an expression. Remember, $-a=-1\cdot{}a.$

Example 7.27

Solution

1. Rewrite the leading minus as a factor. Since $-(y+5)=(-1)(y+5)$, the negative sign can be distributed to both terms.
2. Multiply each product explicitly. The products are $(-1)\cdot{}y=-y$ and $(-1)\cdot{}5=-5$, so the expression becomes $-y+(-5)$.
3. Combine the distributed terms. Writing both products before combining gives $-y-5$. The figures display the negative sign, the factor $-1$, the two separate products, and the equivalent form $-y+(-5)$.
4. State the result. The simplified expression is $-y-5$.

Takeaway: Treat a leading negative sign as multiplication by $-1$ before distributing it across the parentheses.

Sometimes we need to use the Distributive Property as part of the order of operations. Start by looking at the parentheses. If the expression inside the parentheses cannot be simplified, the next step would be multiply using the distributive property, which removes the parentheses. The next two examples will illustrate this.

Example 7.28

Solution

1. Distribute inside the parentheses. Multiply $2$ by each term: $8-2(x+3)=8-[2(x)+2(3)]$.
2. Multiply each product explicitly. The products are $2\cdot{}x=2x$ and $2\cdot{}3=6$, giving $8-(2x+6)$.
3. Remove the parentheses and combine. Writing both products before subtracting gives $8-2x-6$, and the constant terms then combine as $8-6=2$. The figures show these forms and the equivalent result $-2x+2$.
4. State the result. The simplified expression is $2-2x$, equivalently $-2x+2$.

Takeaway: Distribute through parentheses before combining constants, and remember that subtracting a sum subtracts each of its terms.

Example 7.29

Solution

1. Distribute through the first parentheses. Multiply $4$ by each term inside them: $4(x-8)=4(x)+4(-8)$.
2. Multiply each product explicitly. These products are $4\cdot{}x=4x$ and $4\cdot{}(-8)=-32$, so the expression becomes $4x-32-(x+3)$.
3. Remove the remaining parentheses. The subtraction distributes across the sum, so $4x-32-(x+3)=4x-32-x-3$. The figures display this expanded expression before the terms are combined.
4. Combine like terms. The variable terms give $4x-x=3x$, and the constants give $-32-3=-35$, so the final expression is $3x-35$ as shown in the last figure.
5. State the result. The simplified expression is $3x-35$.

Takeaway: Distribute each factor over every term in its parentheses before attempting to combine like terms.

Evaluate Expressions Using the Distributive Property

Some students need to be convinced that the Distributive Property always works.

In the examples below, we will practice evaluating some of the expressions from previous examples; in part ⓐ , we will evaluate the form with parentheses, and in part ⓑ we will evaluate the form we got after distributing. If we evaluate both expressions correctly, this will show that they are indeed equal.

Example 7.30

Solution

1. Substitute $y=10$. Part ⓐ becomes $6(5\cdot{}10+1)=6(50+1)=6(51)$, while part ⓑ becomes $6\cdot{}5(10)+6\cdot{}1=300+6$.
2. Evaluate part ⓐ. Multiplying gives $6\cdot{}51=306$, so the grouped expression equals $306$. The first figures show the substitution into $6(5\cdot{}10+1)$.
3. Evaluate part ⓑ. The separate products are $6\cdot{}5\cdot{}10=300$ and $6\cdot{}1=6$, and $300+6=306$, as shown in the final figures.
4. Compare the results. Part ⓐ and part ⓑ both equal $306$, confirming that distribution preserves the value when $y=10$.

Takeaway: A distributed expression and its original grouped form must agree for every value of the variable, not just for one example.

Example 7.31

Solution

1. Substitute $y=3$. Part ⓐ becomes $-2(4\cdot{}3+1)=-2(12+1)=-2(13)$, while part ⓑ becomes $(-2)\cdot{}4(3)+(-2)\cdot{}1$.
2. Evaluate part ⓐ. Multiplying gives $(-2)\cdot{}13=-26$, so the grouped expression equals $-26$. The figures show the substitution $y=3$ in $-2(4\cdot{}3+1)$.
3. Evaluate part ⓑ. The separate products are $(-2)\cdot{}4\cdot{}3=-24$ and $(-2)\cdot{}1=-2$, so their sum is $-24-2=-26$.
4. Compare the results. Part ⓐ and part ⓑ both equal $-26$, confirming that the negative factor was distributed correctly.

Takeaway: In a negative distributed product, multiply the numerical factors first and keep the negative sign in front of the product.

Example 7.32

Solution

1. Substitute $y=35$ into part ⓐ. The expression becomes $-(35+5)=-(40)$, which equals $-40$. The figures show $35$ replacing $y$ inside the parentheses and the corresponding subtraction form.
2. Substitute $y=35$ into part ⓑ. The expression becomes $-35-5$, and subtracting $5$ from $-35$ gives $-40$.
3. Compare the two values. Part ⓐ equals $-40$ and part ⓑ equals $-40$, so the computations demonstrate that $-(y+5)=-y-5$ for $y=35$.
4. State the result. Both forms evaluate to $-40$.

Takeaway: Distributing $-1$ across a sum negates every term, so $-(a+b)$ is equivalent to $-a-b$.

Section 7.3 Exercises

Section 7.3 Exercises

Simplify Expressions Using the Distributive Property

In the following exercises, simplify using the distributive property.

Evaluate Expressions Using the Distributive Property

In the following exercises, evaluate both expressions for the given value.

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

1. Simplify: $4(x+2).$

2. Simplify: $6(x+7).$

3. Simplify: $9(3y+8).$

4. Simplify: $5(5w+9).$

5. Simplify: $7(x-6).$

1. Simplify: $8(x-5).$

2. Simplify: $\frac{2}{5}(p+10).$

3. Simplify: $\frac{3}{7}(u+21).$

4. Simplify: $6(\frac{5}{6}y+\frac{1}{2}).$

5. Simplify: $12(\frac{1}{3}n+\frac{3}{4}).$

1. Simplify: $100(0.7+0.15p).$

2. Simplify: $100(0.04+0.35d).$

3. Simplify: $r(s-2).$

4. Simplify: $y(z-8).$

5. Simplify: $(x+2)p.$

1. Simplify: $(y+4)q.$

2. Simplify: $-3(6m+5).$

3. Simplify: $-6(8n+11).$

4. Simplify: $-5(2-3a).$

5. Simplify: $-7(8-15y).$

1. Simplify: $-(z-11).$

2. Simplify: $-(x-4).$

3. Simplify: $9-3(x+2).$

4. Simplify: $7x-5(x+4).$

5. Simplify: $6(x-9)-(x+12).$

1. Simplify: $8(x-1)-(x+5).$

2. Evaluate when $w=3:\,$ⓐ $\,5(5w+9)\,$ⓑ $\,5\cdot{}5w+5\cdot{}9.$

3. Evaluate when $y=2:\,$ⓐ $\,9(3y+8)\,$ⓑ $\,9\cdot{}3y+9\cdot{}8.$

4. Evaluate when $n=-2:\,$ⓐ $\,-6(8n+11)$ⓑ $\,-6\cdot{}8n+(-6)\cdot{}11.$

5. Evaluate when $m=-1:$ⓐ $\,\,-3(6m+5)\,$ⓑ $\,-3\cdot{}6m+(-3)\cdot{}5.$

1. Evaluate when $x=36:$ⓐ $\,\,-(x-4)$ⓑ $\,-x+4$ to show that $-(x-4)=-x+4.$

2. Evaluate when $z=55:$ⓐ $\,\,-(z-10)$ⓑ $\,-z+10$ to show that $-(z-10)=-z+10.$

3. $3(a+9)$

4. $9(3w+7)$

5. $7(y-13)$

1. $5(7u-4)$

2. $\frac{1}{3}(u+9)$

3. $\frac{1}{5}(4m+20)$

4. $10(\frac{3}{10}x-\frac{2}{5})$

5. $12(\frac{1}{6}+\frac{3}{4}s)$

1. $u(v-10)$

2. $(a+7)x$

3. $-3(a+11)$

4. $-9(9a+4)$

5. $-4(q-7)$

1. $-6(7x-8)$

2. $-(q+11)$

3. $-(5p-4)$

4. $12+8(u-1)$

5. $18-4(x+2)$

1. $9-6(7n-5)$

2. $8-(r-7)$

3. $-4-(c-10)$

4. $(4y-1)-(y-2)$

5. $9(5u+8)+2(u-6)$

1. $4(6x-1)-(-8)$

2. $11(n-7)-5(n-1)$

3. $7(3n+9)-(4n-13)$

4. If $u=-1,$ evaluate

1. ⓐ $\,8(5u+12)$

2. ⓑ $\,8\cdot{}5u+8\cdot{}12$

5. If $y=\frac{3}{4},$ evaluate

1. ⓐ $\,4(y+\frac{3}{8})$

2. ⓑ $\,4\cdot{}y+4\cdot\frac{3}{8}$

1. If $p=\frac{23}{30},$ evaluate

1. ⓐ $\,-6(5p+11)$

2. ⓑ $\,-6\cdot{}5p+(-6)\cdot{}11$

2. If $n=0.75,$ evaluate

1. ⓐ $\,-100(5n+1.5)$

2. ⓑ $\,-100\cdot{}5n+(-100)(1.5)$

3. If $w=-80,$ evaluate

1. ⓐ $\,-(w-80)$

2. ⓑ $\,-w+80$

4. If $q=0.55,$ evaluate

1. ⓐ $\,-(q+0.48)$

2. ⓑ $\,-q-0.48$

5. Multi-pack purchase Adele’s shampoo sells for $\text{\textdollar 3.97}$ per bottle at the drug store. At the warehouse store, the same shampoo is sold as a $\mathrm{3-pack}$ for $\text{\textdollar 10.49}.$

1. ⓐ Show how you can use the distributive property to find the cost of $3$ bottles bought individually at the drug store.

2. ⓑ How much would Adele save by buying the $\mathrm{3-pack}$ at the warehouse store?

1. Explain how you can multiply $4(\text{\textdollar 5.97})$ without paper or a calculator by thinking of $\text{\textdollar 5.97}$ as $6-0.03$ and then using the distributive property.