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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
\(x^2y^2\) \(1\) \(x^2y^2\)
\(6xy\) \(6\) \(xy\)
\(-3\) \(-3\)

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.

Example13

List all coefficients and variable parts of each term: \(10a^2-5ab-b^2\text{.}\)

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*}

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.

Example14

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.

Example15

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*}
Example16

Simplify \(a^2b^2-ab-2(2a^2b^2-5ab+1)\text{.}\)

Solution

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.

Example17

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. \(\begin{aligned}y^2-y-6 \amp= (-4)^2-(-4)-6\\ \amp =16+4-6\\ \amp=14. \end{aligned}\)

Often algebraic expressions will involve more than one variable.

Example18

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*}
Example19

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

Solution
\begin{equation*} \begin{aligned} \frac{x^2-y^2}{2x-1} \amp = \frac{\left(\alert{-\frac{3}{2}}\right)^2-(\alert{-3})^2}{2\left(\alert{-\frac{3}{2}}\right)-1}\\ \amp = \frac{\frac{9}{4}-9}{-3-1} \end{aligned} \end{equation*}

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.

Example20

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. \(\begin{aligned} \sqrt{b^2-4ac} \amp = \sqrt{(\alert{-7})-4(\alert{-1})\left(\alert{\frac{1}{4}}\right)}\\ \amp = \sqrt{49+4\left(\frac{1}{4}\right)}\\ \amp = \sqrt{49+1}\\ \amp = \sqrt{50}\\ \amp = \sqrt{25\cdot 2}\\ \amp = 5\sqrt{2} \end{aligned}\)

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.

Example21

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.

Example22

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 percent. Percent literally means "per one hundred." We translate \(1\)%, for instance, as "one per one hundred" or we say "one one-hundredth."

SubsectionPercents

Percents are often used to represent the change in some quantity. When representing change, percents give you the ability to see how big the change is relative to the quantity that is changing.

Percent Change

\(\text{percent change (as a decimal)} =\frac{\text{amount of change}}{\text{original amount}}\)

Example23

Your favorite store is having a sale on jeans. Their original price is \($40\text{,}\) but today, their sale price is \($28\text{.}\) What is the percent discount on the jeans?

Solution

The amount of change in the price of the jeans is \(28-40 = -12\) dollars. So the percent change \(= -12/40 = -.3\text{.}\) We convert this to a percent, or \(-30%\text{.}\) The negative tells us that the price has decreased. So the percent discount on the jeans was \(30%\text{.}\)

Example24

The population of Mongolia in 2007 was estimated to be \(2.62\) million. In 2010, it was estimated to be \(2.76\) million. What was the percent increase in the population of Mongolia over this three year period?

Solution

The amount of change in population was \(2.76 - 2.62 = 0.14\) million. So the percent change \(= 0.14/2.76 = 0.051\) (approximate). We convert this to a percent, or \(5.1%\text{.}\) So the population of Mongolia increased by \(5.1%\) between 2007 and 2010.

When working with percent problems, we will often have to change a percentage to a decimal value:

Percent to Decimal Transformation

Suppose \(a\) represents a partial amount of the whole amount \(b\text{.}\) Then we have

\begin{equation*} \frac{\text{partial amount }a}{\text{whole amount }b} = \text{decimal value } \end{equation*}

The decimal value is then converted to a percent.

For example, say we have 15 red marbles in a bag of 75. If we want to know the percentage, we set up the following:

\begin{equation*} \begin{aligned} \frac{\text{partial amount }a}{\text{whole amount } b}\amp = \frac{15}{75} \amp\text{Let }a=15,\, b=75\\ \amp =\frac{15}{75}\cdot\frac{\frac{4}{3}}{\frac{4}{3}}\\ \amp=\frac{20}{100} \\ \amp=0.2\amp\text{Write as a decimal}\\ \amp=20\text{%}\amp\text{Convert to a percent} \end{aligned} \end{equation*}
Example25

Lara's choir consists of 25 singers and 18 are females. What is the percent of females in the choir?

Solution
\begin{equation*} \begin{aligned} \frac{\text{partial amount }a}{\text{whole amount } b}\amp = \frac{18}{25} \amp\text{Let }a=32,\, b=50\\ \amp =\frac{18}{25}\left(\alert{\frac{4}{4}}\right)\\ \amp =\frac{72}{100}\\ \amp =0.72 \amp\text{Write as a decimal}\\ \amp =72\text{%}\amp\text{Convert to a percent} \end{aligned} \end{equation*}

Lara's choir is \(72\)% female.

Areeba's mathematics grade includes 300 points for exams. If the entire amount of points she can earn is 750, what percentage of her grade comes from exams?

Jefferson has 549 followers on Facebook, 215 of which are friends from high school. What percentage of Jefferson's Facebook followers are friends from high school?

Now we can take our knowledge of percent and apply it to banking situations. For example, simple interest \(I\) is given by the formula \(I=prt\text{,}\) where \(p\) represents the principal amount invested at an annual interest rate \(r\) for \(t\) years.

Example28

Calculate the simple interest earned on a \(2\)-year investment of \($1250\) at an annual interest rate of \(3.75\)%.

Solution

Convert \(3.75\)% to a decimal number before using it in the formula:

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

Use this and the fact that \(p=$1250\) and \(t=2\) years to calculate the simple interest:

\begin{equation*} \begin{aligned} I \amp = prt\\ \amp = (\alert{1250})(\alert{0.0375})(\alert{2})\\ \amp = 93.75 \end{aligned} \end{equation*}

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