### Learning Objectives

In this section students will:

Elaina, Ira, and Halle are visiting New York City, and all three will use the subway to get around. Elaina plans to take 20 rides on the subway. Since a new MetroCard costs $1, and each ride costs$2.75, the equation below shows that Elaina can expect to pay $56. $\underbrace{2.75}_{\substack{\text{cost}\\\text{per ride}}}\times\underbrace{20}_{\substack{\text{\# of}\\\text{rides}}}+\underbrace{1}_{\substack{\text{MetroCard}\\\text{cost}}}=56$ If Ira plans to take 15 rides on the subway, the equation $2.75\times 15+1=42.25$ shows that he can expect to pay$42.25. Similarly, Halle’s 12 rides will cost her $2.75\times 12+1=34$ dollars.

At this point, we’ve done three different calculations, but the only quantity that changed in each was the number of rides. To represent all of these calculations in one line, we can let the letter r represent the number of rides and write
$2.75\times r+1$
This is an example of an algebraic expression, which is a powerful idea in math that allows us to represent many different possible calculations in one line. The expression $2.75r+1$ can represent the cost to ride the subway any number of times, and by replacing r with any number, we can choose to focus on one particular calculation.

Since algebraic expressions represent quantities, we can add, subtract, and multiply them like we can with other quantities. Also, since setting one quantity equal to another creates an equation, we can make equations with algebraic expressions. Algebraic expressions and equations are the building blocks of algebra, so this section will help us grow more familiar with them.

### 1.1.1 – Evaluating Algebraic Expressions

In mathematics, we may see expressions such as $x+5,\; \frac{4}{3}\pi {r}^{3},\;$ or $\sqrt{2{m}^{3}{n}^{2}}.$ In the expression $x+5,$ the number $5$ is called a constant because it does not vary; it always has a value of 5. On the other hand, we call x a variable because its value may change. An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division.

You are probably familiar with exponents like $2^6$ or $\left(-4\right)^3$. Exponents provide a quick way to say that we are multiplying by the same number several times. The base of the exponent says which number we are multiplying by and the exponent says how many times. For instance, $2^6 = 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2$ means that we will multiply $2$ by itself $6$ times. The base is $2$ and the exponent is $7$. We can write variables with exponents in the same way.

$$\begin{array}{cccccc}\hfill {\left(-3\right)}^{5}& =& \left(-3\right)\cdot \left(-3\right)\cdot \left(-3\right)\cdot \left(-3\right)\cdot \left(-3\right)\hfill & \hfill {x}^{5}& =& x\cdot x\cdot x\cdot x\cdot x\hfill \\ \hfill {\left(2\cdot 7\right)}^{3}& =& \left(2\cdot 7\right)\cdot \left(2\cdot 7\right)\cdot \left(2\cdot 7\right)\hfill & \hfill \text{ }{\left(yz\right)}^{3}& =& \left(yz\right)\cdot \left(yz\right)\cdot \left(yz\right)\hfill \end{array}$$

In each case, the exponent tells us how many factors of the base to use, regardless of whether the base consists of constants or variables.

Another way we can combine constants and variables is using coefficients. A coefficient is a number written in front of a variable to indicate how many copies of the variable we have. For instance, we might write $3x = x + x + x$ to indicate that we have $3$ copies of $x$. In the expression $3x$, we would say that $3$ is the coefficient of $x$. Likewise, in the expression $5x^2$, we say that $5$ is the coefficient of $x^2$.

Any variable in an algebraic expression may be assigned different values. We could decide that $x = 3$ or we could decide that $x = 8.25$. When we assign a value to $x$ like this, the value of the entire algebraic expression changes. For instance, if we have the expression $5x^2$ and we decide that $x = 3$, then the value of the expression becomes $5(3)^2 = 45$. We say that we evaluated the expression $5x^2$ at $x = 3$ Said otherwise, we evaluate an algebraic expression by plugging in particular values for each of the variables and then simplifying the resulting expression using the order of operations.

### Example 1 – Describing Algebraic Expressions

List the constants and variables for each algebraic expression.

1. x + 5
2. $\frac{4}{3}\pi {r}^{3}$
3. $\sqrt{2{m}^{3}{n}^{2}}$
Constants Variables
a. x + 5 5 x
b. $\frac{4}{3}\pi {r}^{3}$ $\frac{4}{3},\pi$ $r$
c. $\sqrt{2{m}^{3}{n}^{2}}$ 2 $m,n$

### Try It

List the constants and variables for each algebraic expression.

1. $2\pi r\left(r+h\right)$
2. 2(L + W)
3. $4{y}^{3}+y$
Constants Variables
a.$2\pi r\left(r+h\right)$ $2,\pi$ $r,h$
b. 2(L + W) 2 L, W
c. $4{y}^{3}+y$ 4 $y$

### Example 2 – Evaluating an Algebraic Expression at Different Values

Evaluate the expression $2x-7$ for each value for x.

1. $x=0$
2. $x=1$
3. $x=\frac{1}{2}$
4. $x=-4$
1. Substitute 0 for $x.$
$\begin{array}{ccc}\hfill 2x-7& =& 2\left(0\right)-7\\ & =& 0-7\hfill \\ & =& -7\hfill \end{array}$
2. Substitute 1 for $x.$
$\begin{array}{ccc}2x-7& =& 2\left(1\right)-7\hfill \\ & =& 2-7\hfill \\ & =& -5\hfill \end{array}$
3. Substitute $\frac{1}{2}$ for $x.$
$\begin{array}{ccc}\hfill 2x-7& =& 2\left(\frac{1}{2}\right)-7\hfill \\ & =& 1-7\hfill \\ & =& -6\hfill \end{array}$
4. Substitute $-4$ for $x.$
$\begin{array}{ccc}\hfill 2x-7& =& 2\left(-4\right)-7\\ & =& -8-7\hfill \\ & =& -15\hfill \end{array}$

### Try It

Evaluate the expression $11-3y$ for each value for y.

1. $y=2$
2. $y=0$
3. $y=\frac{2}{3}$
4. $y=-5$
1. 5;
2. 11;
3. 9;
4. 26

### Example 3 – Evaluating Algebraic Expressions

Evaluate each expression for the given values.

1. $x+5$ for $x=-5$
2. $\frac{t}{2t-1}$ for $t=10$
3. $\frac{4}{3}\pi {r}^{3}$ for $r=5$
4. $a+ab+b$ for$a=11,b=-8$
5. $\sqrt{2{m}^{3}{n}^{2}}$ for $m=2,n=3$
1. Substitute $-5$ for $x.$
$\begin{array}{ccc}\hfill x+5& =& \left(-5\right)+5\hfill \\ & =& 0\hfill \end{array}$
2. Substitute 10 for $t.$
$\begin{array}{ccc}\hfill \frac{t}{2t-1}& =& \frac{\left(10\right)}{2\left(10\right)-1}\hfill \\ & =& \frac{10}{20-1}\hfill \\ & =& \frac{10}{19}\hfill \end{array}$
3. Substitute 5 for $r.$
$\begin{array}{ccc}\hfill \frac{4}{3}\pi {r}^{3}& =& \frac{4}{3}\pi {\left(5\right)}^{3}\\ & =& \frac{4}{3}\pi \left(125\right)\hfill \\ & =& \frac{500}{3}\pi \hfill \end{array}$
4. Substitute 11 for $a$ and –8 for $b.$
$\begin{array}{ccc}\hfill a+ab+b& =& \left(11\right)+\left(11\right)\left(-8\right)+\left(-8\right)\\ & =& 11-88-8\hfill \\ & =& -85\hfill \end{array}$
5. Substitute 2 for $m$ and 3 for $n.$
$\begin{array}{ccc}\hfill \sqrt{2{m}^{3}{n}^{2}}& =& \sqrt{2{\left(2\right)}^{3}{\left(3\right)}^{2}}\hfill \\ & =& \sqrt{2\left(8\right)\left(9\right)}\hfill \\ & =& \sqrt{144}\hfill \\ & =& 12\hfill \end{array}$

### Try It

Evaluate each expression for the given values.

1. $\frac{y+3}{y-3}$ for $y=5$
2. $7-2t$ for $t=-2$
3. $\frac{1}{3}\pi {r}^{2}$ for $r=11$
4. ${\left({p}^{2}q\right)}^{3}$ for $p=-2,q=3$
5. $4\left(m-n\right)-5\left(n-m\right)$ for $m=\frac{2}{3},n=\frac{1}{3}$
1. 4;
2. 11;
3. $\frac{121}{3}\pi$;
4. 1728;
5. 3

### 1.1.2 – Adding and Subtracting Algebraic Expressions

We can add and subtract algebraic expressions by combining like terms, which are terms that contain the same variables raised to the same exponents. For example, $5{x}^{2}$ and $-2{x}^{2}$ are like terms, and can be added to get $3{x}^{2},$but $3x$ and $3{x}^{2}$ are not like terms, and therefore cannot be added.

### How To

Given multiple algebraic expressions, add or subtract them to simplify the expressions.

1. Combine like terms.
2. Simplify the expression.

### Example 4 – Adding Algebraic Expressions

Find the sum.

$\left(12{x}^{2}+9x-21\right)+\left(4{x}^{3}+8{x}^{2}-5x+20\right)$

$\begin{array}{cc}4{x}^{3}+\left(12{x}^{2}+8{x}^{2}\right)+\left(9x-5x\right)+\left(-21+20\right) \hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Combine like terms}.\hfill \\ 4{x}^{3}+20{x}^{2}+4x-1\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Simplify}.\hfill \end{array}$

#### Analysis

To check your answer for this type of problem, try plugging in different values for the variable $x$. We should get the same value regardless of which expression we evaluate: the original or the simplified one. For instance, if we evaluate both expressions above at $x = 1$, we get

original expression: $(12\cdot 1^2 + 9\cdot 1 - 21) + (4\cdot 1^3 + 8\cdot 1^2 - 5\cdot 1 + 20) = 27$
simplified expression: $4\cdot 1^3 + 20\cdot 1^2 + 4\cdot 1 - 1 = 27$

Try evaluating each of these expressions at other values like $x = 0$ or $x = -2$.

### Try It

Find the sum.

$\left(2{x}^{3}+5{x}^{2}-x+1\right)+\left(2{x}^{2}-3x-4\right)$

$2{x}^{3}+7{x}^{2}-4x-3$

### Example 5 – Subtracting Algebraic Expressions

Find the difference.

$\left(7{x}^{4}-{x}^{2}+6x+1\right)-\left(5{x}^{3}-2{x}^{2}+3x+2\right)$

$\begin{array}{cc}7{x}^{4}-5{x}^{3}+\left(-{x}^{2}+2{x}^{2}\right)+\left(6x-3x\right)+\left(1-2\right)\text{ }\hfill & \phantom{\rule{1em}{0ex}}\text{Combine like terms}.\hfill \\ 7{x}^{4}-5{x}^{3}+{x}^{2}+3x-1\hfill & \phantom{\rule{1em}{0ex}}\text{Simplify}.\hfill \end{array}$

#### Analysis

Note that finding the difference between two algebraic expressions is the same as adding the opposite of the second algebraic expression to the first.

### Try It

Find the difference.

$\left(-7{x}^{3}-7{x}^{2}+6x-2\right)-\left(4{x}^{3}-6{x}^{2}-x+7\right)$

$-11{x}^{3}-{x}^{2}+7x-9$

### 1.1.3 – Multiplying Algebraic Expressions

Multiplying algebraic expressions is a bit more challenging than adding and subtracting algebraic expressions. We must use the distributive property to multiply each term in the first algebraic expression by each term in the second algebraic expression. We then combine like terms. We can also use a shortcut called the FOIL method when multiplying algebraic expressions with two terms.

#### Multiplying Algebraic Expressions Using the Distributive Property

To multiply a number by an algebraic expression, we use the distributive property. The number must be distributed to each term of the algebraic expression. We can distribute the $2$ in $2\left(x+7\right)$ to obtain the equivalent expression $2x+14.$ When multiplying algebraic expressions, the distributive property allows us to multiply each term of the first algebraic expression by each term of the second. We then add the products together and combine like terms to simplify.

### How To

Given the multiplication of two algebraic expressions, use the distributive property to simplify the expression.

1. Multiply each term of the first algebraic expression by each term of the second.
2. Combine like terms.
3. Simplify.

### Example 6 – Multiplying Algebraic Expressions Using the Distributive Property

Find the product.

$\left(2x+1\right)\left(3{x}^{2}-x+4\right)$

$\begin{array}{cc}2x\left(3{x}^{2}-x+4\right)+1\left(3{x}^{2}-x+4\right) \hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Use the distributive property}.\hfill \\ \left(6{x}^{3}-2{x}^{2}+8x\right)+\left(3{x}^{2}-x+4\right)\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Multiply}.\hfill \\ 6{x}^{3}+\left(-2{x}^{2}+3{x}^{2}\right)+\left(8x-x\right)+4\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Combine like terms}.\hfill \\ 6{x}^{3}+{x}^{2}+7x+4 \hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Simplify}.\hfill \end{array}$

#### Analysis

We can use a table to keep track of our work, as shown below. Write one algebraic expression across the top and the other down the side. For each box in the table, multiply the term for that row by the term for that column. Then add all of the terms together, combine like terms, and simplify.

 $3{x}^{2}$ $-x$ $+4$ $2x$ $6{x}^{3}$ $-2{x}^{2}$ $8x$ $+1$ $3{x}^{2}$ $-x$ $4$

### Try It

Find the product.

$\left(3x+2\right)\left({x}^{3}-4{x}^{2}+7\right)$

$3{x}^{4}-10{x}^{3}-8{x}^{2}+21x+14$

#### Using FOIL to Multiply Algebraic Expressions with Two Terms

A shortcut called FOIL is sometimes used to find the product of two algebraic expressions with two terms. It is called FOIL because we multiply the first terms, the outer terms, the inner terms, and then the last terms of each algebraic expression.

The FOIL method arises out of the distributive property. We are simply multiplying each term of the first algebraic expression by each term of the second algebraic expression, and then combining like terms.

### How To

Given two algebraic expressions with two terms, use FOIL to simplify the expression.

1. Multiply the first terms of each algebraic expression.
2. Multiply the outer terms of the algebraic expressions.
3. Multiply the inner terms of the algebraic expressions.
4. Multiply the last terms of each algebraic expression.
6. Combine like terms and simplify.

### Example 7 – Using FOIL to Multiply Algebraic Expressions with Two Terms

Use FOIL to find the product.

$\left(2x-18\right)\left(3x+3\right)$

Find the product of the first terms.

Find the product of the outer terms.

Find the product of the inner terms.

Find the product of the last terms.

$\begin{array}{cc}6{x}^{2}+6x-54x-54\hfill & \phantom{\rule{2em}{0ex}}\text{Add the products}.\hfill \\ 6{x}^{2}+\left(6x-54x\right)-54\hfill & \phantom{\rule{2em}{0ex}}\text{Combine like terms}.\hfill \\ 6{x}^{2}-48x-54\hfill & \phantom{\rule{2em}{0ex}}\text{Simplify}.\hfill \end{array}$

### Try It

Use FOIL to find the product.

$\left(x+7\right)\left(3x-5\right)$

$3{x}^{2}+16x-35$

### 1.1.4 – Performing Operations with Algebraic Expressions of Several Variables

We have looked at algebraic expressions containing only one variable. However, an algebraic expression can contain several variables. All of the same rules apply when working with algebraic expressions containing several variables. Consider an example:

$\begin{array}{cc}\left(a+2b\right)\left(4a-b-c\right)\hfill & \hfill \\ a\left(4a-b-c\right)+2b\left(4a-b-c\right)\hfill & \phantom{\rule{2em}{0ex}}\text{Use the distributive property}.\hfill \\ 4{a}^{2}-ab-ac+8ab-2{b}^{2}-2bc\hfill & \phantom{\rule{2em}{0ex}}\text{Multiply}.\hfill \\ 4{a}^{2}+\left(-ab+8ab\right)-ac-2{b}^{2}-2bc\hfill & \phantom{\rule{2em}{0ex}}\text{Combine like terms}.\hfill \\ 4{a}^{2}+7ab-ac-2bc-2{b}^{2}\hfill & \phantom{\rule{2em}{0ex}}\text{Simplify}.\hfill \end{array}$

### Example 8 – Multiplying Algebraic Expressions Containing Several Variables

Multiply $\left(x+4\right)\left(3x-2y+5\right).$

Follow the same steps that we used to multiply algebraic expressions containing only one variable.

$\begin{array}{cc}x\left(3x-2y+5\right)+4\left(3x-2y+5\right) \hfill & \phantom{\rule{2em}{0ex}}\text{Use the distributive property}.\hfill \\ 3{x}^{2}-2xy+5x+12x-8y+20\hfill & \phantom{\rule{2em}{0ex}}\text{Multiply}.\hfill \\ 3{x}^{2}-2xy+\left(5x+12x\right)-8y+20\hfill & \phantom{\rule{2em}{0ex}}\text{Combine like terms}.\hfill \\ 3{x}^{2}-2xy+17x-8y+20 \hfill & \phantom{\rule{2em}{0ex}}\text{Simplify}.\hfill \end{array}$

### Try It

Multiply $\left(3x-1\right)\left(2x+7y-9\right).$

$6{x}^{2}+21xy-29x-7y+9$

### 1.1.5 – Testing Possible Solutions to Algebraic Equations

The algebraic expression $2.75r+1$ we saw at the beginning of this section represented the cost to take $r$ rides on the subway. Now, consider the following statement:

The cost to take $r$ rides on the subway is $67$ dollars

In the language of math, we can write the same statement as

$2.75r+1=67$

This is an example of an algebraic equation, which is very different from an algebraic expression. An algebraic equation is a statement that two algebraic expressions are equal: the expressions to the left and right of the equal sign. The equation above simply states that the expression $2.75r+1$ is equal to the expression $67$.

A solution to an equation is a value for the variable (or variables if there are more than one) that makes the equation true upon evaluation. For example, $r=0$ is not a solution to the equation $2.75r+1=67$ because when we substitute $0$ for $r$, we get the equation $1=67$ which is certainly not true. However, you can check that $r=24$ is a solution to this equation. Later on in the course we’ll learn how to find solutions for many different types of equations, but for now it is important to understand the difference between an algebraic expression and an algebraic equation, and how to test possible solutions to equations.

### Example 9 – Testing Possible Solutions to Algebraic Equations

Determine if the given value(s) for the variable(s) represents a solution to the equation.

1. $2(p+6)-1=7(p+3)$, $p=2$
2. $2(p+6)-1=7(p+3)$, $p=-2$
3. $d=d+1$, $d=0$
4. $y=-6x+9$, $x=3$, $y=-9$
5. $x^2=2x+3$, $x=-1$

We can evaluate the algebraic expressions on the left and right at the given values to see if the resulting equation is true.

1. $2(p+6)-1=7(p+3)$, $p=2$
$\begin{array}{rcl} 2(p+6)-1&=&7(p+3)\\ 2(2+6)-1&\stackrel{?}{=}&7(2+3)\\ 2(8)-1&\stackrel{?}{=}&7(5)\\ 16-1&\stackrel{?}{=}&35\\ 15&\ne& 35 \end{array}$Since 15 is not equal to 35, we can say that $p=2$ is not a solution to the equation. Notice that we write a question mark over the equal sign to test a possible solution. This simply indicates that we don’t yet know if the equation is true until we finish evaluating.
2. $2(p+6)-1=7(p+3)$, $p=-2$
$\begin{array}{rcl} 2(p+6)-1&=&7(p+3)\\ 2(-2+6)-1&\stackrel{?}{=}&7(-2+3)\\ 2(4)-1&\stackrel{?}{=}&7(1)\\ 8-1&\stackrel{?}{=}&7\\ 7&=& 7 \end{array}$So $p=-2$ is a solution to the equation.
3. $d=d+1$, $d=0$
$\begin{array}{rcl} d&=&d+1\\ 0&\stackrel{?}{=}&0+1\\ 0&\ne& 1 \end{array}$So $d=0$ is not a solution to the equation. (Does this equation have any solution?)
4. $y=-6x+9$, $x=3$, $y=-9$This equation has two variables, so we substitute the given values for each to test the possible solution.$\begin{array}{rcl} y&=&-6x+9\\ -9&\stackrel{?}{=}&-6(3)+9\\ -9&\stackrel{?}{=}&-18+9\\ -9&=& -9 \end{array}$So $x=3$ and $y=-9$ is a solution to the equation. (Can you find any other solutions?)
5. $x^2=2x+3$, $x=3$
$\begin{array}{rcl} x^2&\stackrel{?}{=}&2x+3\\ (-1)^2&\stackrel{?}{=}&2(-1)+3\\ 1&\stackrel{?}{=}&-2+3\\ 1&=& 1 \end{array}$So $x=-1$ is a solution to the equation. (Try $x=3$ for another solution.)

### Try It

Determine if the given value(s) for the variable(s) represents a solution to the equation.

1. $6(j-6)+8=-4(j-9)-84$, $j=2$
2. $H^2=H$, $H=-1$
3. $y=8x-2$, $x=-1$, $y=-6$
4. $x^2=3x+28$, $x=4$

None of these are solutions, but if you switch the sign of each given value, then they all become solutions!

### 1.1.6 – Translate Words into Algebraic Expressions

We started this section with an example about subway fares. Recall that Elaina, Ira, and Halle each performed different calculations to determine their subway fares, but the only quantity that changed across calculations was the number of rides. We thus made $r$ a variable representing the number of rides, and came up with the following expression to represent the cost of riding the subway.

$2.75 \times r + 1$

To really make use of mathematics, we must learn how to translate real-life situations from words into algebraic expressions. Just as in the subway example, we can start by performing some sample calculations and then ask ourselves what quantities changed in these calculations? Anything that changes will need to be written as a variable

### How To

Find an expression that represents a real-life situation.

1. Perform some sample calculations.
2. Determine which quantities change from one calculation to the other, and create a variable to represent each of these quantities.
3. Replace the quantities in your sample calculations with the corresponding variable.

### Example 10 – Finding an expression to describe a real-life situation

Jazmin and Devon go to the deli to get snacks for their friends. Jazmin wants to buy three bags of chips and two cans of soda. Devon plans to get only one bag of chips, but he plans to buy four cans of soda. A bag of chips costs $1.50, and a can of soda costs$2. Write an expression describing the cost of the snacks.

1. First, let’s perform some sample calculations. Jazmin will pay $1.50\cdot 3 + 2\cdot 2 = \8.50$ for her snacks. Devon will pay $1.50\cdot 1 + 2\cdot 4 = \9.50$.
2. What changes in each of these calculations? The prices are always the same, $1.50 for chips and$2 for soda, so these numbers should remain as constants. On the other hand, the quantity of chips and soda purchased differ for Jazmin and Devon. We need to represent each of these quantities as a variable. Let $c$ represent how many bags of chips are purchased and let $s$ reprsent how many cans of soda are purchased.
3. Finally, we replace each quantity in our sample calculation with the corresponding variable. For instance, Devon calculated $1.50\cdot 1 + 2\cdot 4$ for 1 bag of chips ($c$) and 4 cans of soda ($s$]). We replace these quantities with the variables $c$ and $s$ to get the expression $1.50c + 2s$

### Try It

Ghalib rides the bus to school. It takes him 5 minutes to walk to the bus stop; there are 12 stops on the bus route before he gets to school. Cristina also rides the bus to school. It takes her 20 minutes to walk to the bus stop, but she only has 4 stops on her route. Suppose the bus always comes right away (yeah right!) and it takes the bus 3 minutes to travel from one stop to the next. Write an expression that describes how long it should take to get to school.

First, we perform some sample calculations. It will take Ghalib $5 + 3\cdot 12 = 41$ minutes to get to school. It will take Cristina $20 + 3\cdot 4 = 32$ minutes to get to school. Notice that in both calculations, the number of minutes per stop doesn't change. On the other hand, the length of the walk to the bus stop and the number of stops on the route do change. We need variables that represent each of these quantities. Let $w$ represent the length of the walk to the bus stop (in minutes) and let $s$ represent how many bus stops are on the route. Replacing the quantities in our calculations above with these numbers gives us the expression $w + 3s$.

Access these online resources for additional instruction and practice with algebraic expressions.

### Key Concepts

• Algebraic expressions are composed of constants and variables that are combined using addition, subtraction, multiplication, and division (see Example 1). They take on a numerical value when evaluated by replacing variables with constants (see Example 2 and Example 3).
• We can add and subtract algebraic expressions by combining like terms (see Example 4 and Example 5).
• To multiply algebraic expressions, use the distributive property to multiply each term in the first algebraic expression by each term in the second. Then add the products (see Example 6).
• FOIL (First, Outer, Inner, Last) is a shortcut that can be used to multiply algebraic expressions with two terms (see Example 7).
• Follow the same rules to work with algebraic expressions containing several variables (see Example 8).
• Algebraic equations are statements that two expressions are equal: the expressions to the left and right of the equal sign. A solution to an algebraic equation is a value for the variable that makes the equation true upon evaluation (see Example 9).
• To translate real-life situations from words into algebraic expressoins, we start by performing some sample calculations. Then, we identify which quantities change from one calculation to the other and replace these quantities with variables (see Example 10).

### Glossary

constant
a quantity that does not change value
variable
a quantity that may change value
algebraic expression
constants and variables combined using addition, subtraction, multiplication, and division
exponential notation
a shorthand way to say that we are multiplying by the same number several times
base (of an exponent)
in exponential notation, the constant or variable that is being multiplied
exponent
in exponential notation, the raised number or variable that indicates how many times the base is being multiplied
coefficient
a number written in front of a variable to indicate how many copies of the variable are present
evaluate an expression
to plug in a specific number for each variable and simplify the resulting expression
distributive property
the product of a factor times a sum is the sum of the factor times each term in the sum; in symbols, $a\cdot \left(b+c\right)=a\cdot b+a\cdot c$
equation
a mathematical statement indicating that two expressions are equal
solution
a value for a variable (or variables) that makes an equation true upon evaluation