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## SectionAlgebraic Expressions and Formulas

### SubsectionAlgebraic Expressions and the Distributive Property

In algebra, letters called variables are used to represent numbers. Combinations of variables and numbers along with mathematical operations form algebraic expressions, or just expressions. The following are some examples of expressions with one variable, $x\text{:}$

\begin{gather*} 2x+3,~~~ x^2-9,~~~ \frac{1}{x}+\frac{x}{x+2}, ~~~ 3\sqrt{x}+x \end{gather*}

Terms in an algebraic expression are separated by addition operators and factors are separated by multiplication operators. The numerical factor of a term is called the coefficient. For example, the algebraic expression $x^2y^2+6xy-3$ can be thought of as $x^2y^2+6xy+(-3)$ and has three terms. The first term, $x^2y^2\text{,}$ represents the quantity $1x^2y^2=1\cdot x\cdot x\cdot y\cdot y$ where $1$ is the coefficient and $x$ and $y$ are the variables. All of the variable factors with their exponents form the variable part of a term. If a term is written without a variable factor, then it is called a constant term. Below we see the components of $x^2y^2+6xy-3\text{.}$

 Terms Coefficient Variable Part $x^2y^2$ $1$ $x^2y^2$ $6xy$ $6$ $xy$ $-3$ $-3$

Notice that the first term has a coefficient of $1\text{.}$ Since we can always multiple an expression by $1$ without changing its value, we say that a term without a coefficient has a coefficient of $1\text{.}$

The third term in this expression, $-3\text{,}$ is called a constant term because it is written without a variable factor. While a variable represents an unknown quantity and may change, the constant term does not change. Also, notice that the constant term is $-3\text{,}$ not just $3\text{.}$ Since terms are defined by addition, whenever we see subtraction, we can think of it as making the coefficient negative.

###### Example19

List all coefficients and variable parts of each term in the expression $10a^2-5ab-b^2\text{.}$

Solution

We want to think of the third term in this example, $-b^2\text{,}$ as $-1b^2.$

 Terms Coefficient Variable Part $10a^2$ $10$ $a^2$ $-5ab$ $-5$ $ab$ $-b^2$ $-1$ $b^2$

In our study of algebra, we will encounter a wide variety of algebraic expressions. Typically, expressions use the two most common variables, $x$ and $y\text{.}$ However, expressions may use any letter (or symbol) for a variable, even Greek letters, such as alpha ($\alpha$) and beta ($\beta$). Some letters and symbols are reserved for special constants, such as $\pi\approx 3.14159\text{.}$

The properties of real numbers are important in our study of algebra because a variable is simply a letter that represents a real number. In particular, the distributive property states that if given any real numbers $a\text{,}$ $b$ and $c\text{,}$ then

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

Since the order does not matter for multiplication, we can apply the distributive property on the other side as well:

\begin{equation*} (b+c)a=ba+ca \end{equation*}

This property is one that we apply often when simplifying algebraic expressions. To demonstrate how it will be used, we simplify $2(5-3)$ in two ways, and observe the same correct result.

\begin{gather*} \text{Working parenthesis first: } 2(5-3)=2(2)=4,\\ \text{Using the distributive property: } 2(5-3)=2\cdot 5-2\cdot 3=10-6=4. \end{gather*}

Certainly, if the contents of the parentheses can be simplified we should do that first. On the other hand, when the contents of parentheses cannot be simplified any further, we multiply every term within it by the factor outside of it using the distributive property. Applying the distributive property allows us to multiply and remove the parentheses.

###### Example20

Simplify $5(-2a+5b)-2c\text{.}$

Solution

Multiply only the terms grouped within the parentheses for which we are applying the distributive property.

\begin{equation*} \begin{aligned} 5(-2a+5b)-2c \amp = 5\cdot (-2a)+5\cdot 5b-2c\\ \amp = -10a+25b-2c. \end{aligned} \end{equation*}

Terms whose variable parts have the same variables with the same exponents are called like terms. Furthermore, constant terms are considered to be like terms. If an algebraic expression contains like terms, we can apply the distributive property as follows:

\begin{equation*} \begin{aligned} 5x+7x \amp = (5+7)x = 12x\\ 4x^2+5x^2-7x^2\amp=(4+5-7)x^2=2x^2. \end{aligned} \end{equation*}

In other words, if the variable parts of terms are exactly the same, then we can add or subtract the coefficients to obtain the coefficient of a single term with the same variable part. This process is called combining like terms. For example,

\begin{equation*} 12x^2y^3+3x^2y^3=15x^2y^3. \end{equation*}

Notice that the variable factors and their exponents do not change. Combining like terms in this manner, so that the expression contains no other similar terms, is called simplifying the expression. Use this idea to simplify algebraic expressions with multiple like terms.

###### Example21

Simplify $x^2-10x+8+5x^2-6x-1\text{.}$

Solution

Identify the like terms and add the corresponding coefficients.

\begin{equation*} \begin{aligned} 1x^2-10x+8+5x^2-6x-1 = 6x^2-16x+7 \end{aligned} \end{equation*}
###### Example22

Simplify $a^2b^2-ab-2(2a^2b^2-7ab+1+2ab)\text{.}$

Solution

Before applying the distributive property, we should try to combine like terms.

\begin{equation*} \begin{aligned} a^2b^2-ab-2(2a^2b^2-7ab+1+2ab)\amp=a^2b^2-ab-2(2a^2b^2-5ab+1) \end{aligned} \end{equation*}

Next we distribute $-2$ and then combine like terms.

\begin{equation*} \begin{aligned} a^2b^2-ab-2(2a^2b^2-5ab+1) \amp = a^2b^2-ab-4a^2b^2+10ab-2\\ \amp = -3a^2b^2+9ab-2 \end{aligned} \end{equation*}

### SubsectionEvaluating Algebraic Expressions

An algebraic expression can be thought of as a generalization of particular arithmetic operations. Performing these operations after substituting given values for variables is called evaluating. In algebra, a variable represents an unknown value. However, if the problem specifically assigns a value to a variable, then you can replace that letter with the given number and evaluate using the order of operations.

###### Example23

Evaluate:

1. $5x-2$ where $x=\frac{2}{3}\text{,}$
2. $y^2-y-6$ where $y=-4\text{.}$
Solution

We replace, or substitute, the given value for the variable.

1. \begin{aligned}5x-2 \amp= 5 \left(\frac{2}{3}\right)-2\\ \amp =\frac{10}{3}-\frac{2}{1}\cdot\frac{3}{3}\\ \amp=\frac{10-6}{3}\\ \amp=\frac{4}{3}, \end{aligned}
2. \begin{aligned}y^2-y-6 \amp= (-4)^2-(-4)-6\\ \amp =16+4-6\\ \amp=14. \end{aligned}
###### Example24

Evaluate:

1. $1+x+x^2+x^3$ where $x=1\text{,}$
2. $1+x+x^2+x^3$ where $x=-1\text{.}$
Solution

We replace, or substitute, the given value for the variable.

1. \begin{aligned}1+x+x^2+x^3 \amp= 1+(1)+(1)^2+(1)^3 \\ \amp=1+1+1+1 \\ \amp=4 \end{aligned}
2. \begin{aligned}1+x+x^2+x^3 \amp= 1+(-1)+(-1)^2+(-1)^3 \\ \amp=1-1+1-1 \\ \amp=0 \end{aligned}

Often algebraic expressions will involve more than one variable.

###### Example25

Evaluate $a^3-8b^3$ where $a=-1$ and $b=\frac{1}{2}\text{.}$

Solution

After substituting in the appropriate values, we must take care to simplify using the correct order of operations.

\begin{equation*} \begin{aligned} a^3-8b^3 \amp = (\alert{-1})^3-8\left(\alert{\frac{1}{2}}\right)^3\\ \amp = -1-8\left(\frac{1}{8}\right)\\ \amp = -1-1\\ \amp = -2 \end{aligned} \end{equation*}
###### Example26

Evaluate $\frac{x^2-y^2}{2x-1}$ where $x=-\frac{3}{2}$ and $y=-3\text{.}$

Solution

At this point we have a complex fraction. Simplify the numerator and then multiply by the reciprocal of the denominator.

\begin{equation*} \begin{aligned} \amp = \frac{\frac{9}{4}-\frac{9}{1}\cdot\alert{\frac{4}{4}}}{-4}\\ \amp = \frac{-\frac{27}{4}}{\frac{-4}{1}}\\ \amp = \frac{-27}{4}\left(-\frac{1}{4}\right)\\ \amp=\frac{27}{16} \end{aligned} \end{equation*}

The answer to the previous example can be written as a mixed number, $\frac{27}{16}=1\frac{11}{16}\text{.}$ Unless the original problem has mixed numbers in it, or it is an answer to a real-world application, solutions in this course will be expressed as reduced improper fractions.

###### Example27

Evaluate $\sqrt{b^2-4ac}$ where $a=-1,\, b=-7$ and $c=\frac{1}{4}\text{.}$

Solution

Substitute in the appropriate values and then simplify. \begin{aligned} \sqrt{b^2-4ac} \amp = \sqrt{(\alert{-7})^2-4(\alert{-1})\left(\alert{\frac{1}{4}}\right)}\\ \amp = \sqrt{49+4\left(\frac{1}{4}\right)}\\ \amp = \sqrt{49+1}\\ \amp = \sqrt{50}\\ \amp = \sqrt{25\cdot 2}\\ \amp = 5\sqrt{2} \end{aligned}

### SubsectionFormulas

The main difference between algebra and arithmetic is the organized use of variables. This idea leads to reusable formulas. A formula is an equation that relates several variables to describe common applications. For example, the equation

\begin{equation*} P = 2l + 2w \end{equation*}

gives the perimeter of a rectangle in terms of its length and width.

Suppose we have some wire fence to enclose an exercise area for rabbits, and we would like to see what dimensions are possible for different rectangles with that perimeter. In this case, it would be more useful to have a formula for the length of the rectangle in terms of its perimeter and its width. We can find such a formula by solving the perimeter formula for $l$ in terms of $P$ and $w\text{.}$

\begin{align*} 2l + 2w \amp= P \amp\amp\text{Subtract }2 w \text{ from both sides}\\ 2l \amp= P - 2w \amp\amp\text{Divide both sides by 2}\\ l \amp= \frac{P - 2w}{2} \end{align*}

The result is a new formula that gives the length of a rectangle in terms of its perimeter and its width.

###### Example28

Solve $3x - 5y = 40$ for $y$ in terms of $x\text{.}$

Solution

We isolate $y$ on one side of the equation.

\begin{align*} 3x - 5y \amp = 40\amp\amp\text{Subtract }3x \text{ from both sides.}\\ -5y \amp = 40 - 3x\amp\amp\text{Divide both sides by }-5.\\ \frac{-5y}{-5} \amp = \frac{40 - 3x}{-5}\amp\amp\text{Simplify both sides.}\\ y \amp = -8 + \frac{3}{5}x \end{align*}

There are many real world applications for which we can use formulas. One such useful formula is that which relates Fahrenheit to Celsius.

###### Example29

The formula $5F = 9C + 160$ relates the temperature in degrees Fahrenheit, $F\text{,}$ to the temperature in degrees Celsius, $C\text{.}$ Solve the formula for $C$ in terms of $F\text{.}$

Solution

We begin by isolating the term that contains $C\text{.}$

\begin{align*} 5F \amp= 9C + 160 \amp\amp\text{Subtract 160 from both sides.}\\ 5F - 160 \amp = 9C \amp\amp\text{Divide both sides by 9.}\\ \frac{5F - 160}{9} \amp =C \end{align*}

We can also write the formula for $C$ in terms of $F$ as $C = \dfrac{5}{9}F - \dfrac{160}{9}\text{.}$

Formulas can be found in a multitude of subjects. One especially useful everyday application is that of percents. For example, simple interest $I$ is given by the formula $I=prt$ where $p$ represents the principal amount invested at an annual interest rate $r$ (as a decimal, not a percent) for $t$ years.

###### Example30

Calculate the simple interest earned on a $2$-year investment of \$$1250$ at an annual interest rate of $3.75\text{.}$

Solution

We must convert $3.75$% to a decimal number before using it in the formula:

\begin{equation*} r=3.75\text{%}=0.0375 \end{equation*}

We also know that $p=1235$ and $t=2\text{,}$ so

The simple interest earned is $93.75\text{.}$