You are familiar with the use of letters, or variables, to stand for unknown numbers in equations or formulas. Variables are also used to represent numerical quantities that change over time or in different situations. For example, \(p\) might stand for the atmospheric pressure at different heights above the Earth's surface. Or \(N\) might represent the number of people infected with cholera \(t\) days after the start of an epidemic.
An algebraic expression is any meaningful combination of numbers, variables, and symbols of operation. Algebraic expressions are used to express relationships between variable quantities.
Example A.17
Loren makes $\(6\) an hour working at the campus bookstore.
Choose a variable for the number of hours Loren works per week.
Write an algebraic expression for the amount of Loren's weekly earnings.
Solution
Let \(h\) stand for the number of hours Loren works per week.

The amount Loren earns is given by
\begin{equation*}
\blert{6\times (\text{number of hours Loren worked})}
\end{equation*}
or \(6\cdot h\text{.}\) Loren's weekly earnings can be expressed as \(6h\text{.}\)
The algebraic expression \(6h\) represents the amount of money Loren earns in terms of the number of hours she works. If we substitute a specific value for the variable in an expression, we find a numerical value for the expression. This is called evaluating the expression.
Example A.18
If Loren from Example A.17 works for \(16\) hours in the bookstore this week, how much will she earn?
Solution
Evaluate the expression \(6h\) for \(h=\alert{16}\text{.}\)
\begin{equation*}
6h = 6(\alert{16}) = 96
\end{equation*}
Loren will make $\(96\text{.}\)
Example A.19
April sells environmentally friendly cleaning products. Her income consists of $\(200\) per week plus a commission of \(9\%\) of her sales.
Choose variables to represent the unknown quantities and write an algebraic expression for April's weekly income in terms of her sales.
Find April's income for a week in which she sells $\(350\) worth of cleaning products.
Solution

Let \(I\) represent April's total income for the week, and let \(S\) represent the total amount of her sales. We translate the information from the problem into mathematical language as follows:
\begin{gather*}
\blert{\text{Her income consists of }$200 . . .\text{ plus }. . . 9\% \text{ of her sales}} \\
I \hphantom{consists of}= \hphantom{of}200 \hphantom{plus+}+ \hphantom{....}0.09 \hphantom{of her}S
\end{gather*}
Thus, \(I = 200 + 0.09S\text{.}\)

We want to evaluate our expression from part (a) with \(S = 350\text{.}\) We substitute \(\alert{350}\) for \(S\) to find
\begin{equation*}
I = 200 + 0.09(\alert{350})
\end{equation*}
Following the order of operations, we perform the multiplication before the addition. Thus, we begin by computing \(0.09(350)\text{.}\)
\begin{align*}
I \amp = 200 + 0.09(350)\amp\amp\blert{\text{Multiply }0.09 (350) \text{ first.}}\\
\amp = 200 + 31.5\\
\amp = 231.50
\end{align*}
April's income for the week is $\(231.50\text{.}\)
Example A.21
Economy Parcel Service charges $\(2.80\) per pound to deliver a package from Pasadena to Cedar Rapids. Andrew wants to mail a painting that weighs \(8.3\) pounds, plus whatever packing material he uses.
Choose variables to represent the unknown quantities and write an expression for the cost of shipping Andrew's painting.
Find the shipping cost if Andrew uses \(2.9\) pounds of packing material.
Solution

Let \(C\) stand for the shipping cost and let \(w\) stand for the weight of the packing material. Andrew must find the total weight of his package first, then multiply by the shipping charge.
The total weight of the package is \(8.3 + w\) pounds. We use parentheses around this expression to show that it should be computed first, and the sum should be multiplied by the shipping charge of $\(2.80\) per pound. Thus,
\begin{equation*}
C = 2.80(8.3 + w)
\end{equation*}

Evaluate the formula from part (a) with \(w = \alert{2.9}\text{.}\)
\begin{align*}
C \amp = 2.80(8.3 + \alert{2.9})\amp\amp\blert{\text{Add inside parentheses.}}\\
\amp = 2.80(11.2)\amp\amp\blert{\text{Multiply.}}\\
\amp = 31.36
\end{align*}
The cost of shipping the painting is $\(31.36\text{.}\)
Subsection Problem Solving
Problem solving often involves translating a reallife problem into a computer programming language, or, in our case, into algebraic expressions. We can then use algebra to solve the mathematical problem and interpret the solution in the context of the original problem. Here are some guidelines for problem solving with algebraic equations.
Guidelines for Problem Solving
Identify the unknown quantity and assign a variable to represent it.
Find some quantity that can be expressed in two different ways and write an equation.
Solve the equation.
Interpret your solution to answer the question in the problem.
In step 1, begin by writing an English phrase to describe the quantity you are looking for. Be as specific as possible—if you are going to write an equation about this quantity, you must understand its properties! Remember that your variable must represent a numerical quantity. For example, \(x\) can represent the speed of a train, but not just “the train.”
Writing an equation is the hardest part of the problem. Note that the quantity mentioned in step 2 will probably not be the same unknown quantity you are looking for, but the algebraic expressions you write will involve your variable. For example, if your variable represents the speed of a train, your equation might be about the distance the train traveled.
Subsection Percent Problems
Recall the basic formula for computing percents.
Percent Formula
\begin{equation*}
P = rW
\end{equation*}
the Part (or percent) = the percentage rate \(\times\) the Whole Amount
A percent increase or percent decrease is calculated as a fraction of the original amount. For example, suppose you make $\(16.00\) an hour now, but next month you are expecting a \(5\%\) raise. Your new salary should be
\begin{equation*}
\stackrel{\text{Original salary}}{\$16.00} + \stackrel{\text{Increase}}{0.05(\$16.00)} = \stackrel{\text{New Salary}}{\$16.80}
\end{equation*}
Example A.25
The price of housing in urban areas increased \(4\%\) over the past year. If a certain house costs $\(100,000\) today, what was its price last year?
Solution
Let \(c\) represent the cost of the house last year.

Express the current price of the house in two different ways. During the past year, the price of the house increased by \(4\%\text{,}\) or \(0.04c\text{.}\) Its current price is thus
\begin{equation*}
\stackrel{\text{Original cost}}{(1)c} + \stackrel{\text{Price increase}}{0.04c} = c(1 + 0.04) = 1.04c
\end{equation*}
This expression is equal to the value given for current price of the house:
\begin{equation*}
1.04c = 100,000
\end{equation*}

To solve this equation, we divide both sides by \(1.04\) to find
\begin{equation*}
c = \frac{100,000}{1.04}= 96,153.846
\end{equation*}
To the nearest cent, the cost of the house last year was $\(96,153.85\text{.}\)
Subsection Weighted Averages
We find the average, or mean, of a set of values by adding up the values and dividing the sum by the number of values. Thus, the average, \(\overline{x}\text{,}\) of the numbers \(x_1, x_2, \ldots , x_n\) is given by
\begin{equation*}
\overline{x} = \frac{x_1 + x_2 + \cdot \cdot \cdot +x_n}{n}
\end{equation*}
In a weighted average, the numbers being averaged occur with different frequencies or are weighted differently in their contribution to the average value. For instance, suppose a biology class of 12 students takes a 10point quiz. Of the 12 students, 2 receive 10s, three receive 9s, 5 receive 8s, and 2 receive scores of 6. The average score earned on the quiz is then
\begin{equation*}
\overline{x} = \frac{\alert{2}(10) + \alert{3}(9) + \alert{5}(8) + \alert{2}(6)}{12}=8.25
\end{equation*}
The numbers in color are called the weights—in this example they represent the number of times each score was counted. Note that \(n\text{,}\) the total number of scores, is equal to the sum of the weights:
\begin{equation*}
12 = 2 + 3 + 5 + 2
\end{equation*}
Example A.27
Kwan's grade in his accounting class will be computed as follows: Tests count for \(50\%\) of the grade, homework counts for \(20\%\text{,}\) and the final exam counts for \(30\%\text{.}\) If Kwan has an average of \(84\) on tests and \(92\) on homework, what score does he need on the final exam to earn a grade of \(90\text{?}\)
Solution
Let \(x\) represent the final exam score Kwan needs.

Kwan's grade is the weighted average of his test, homework, and final exam scores.
\begin{equation*}
\frac{\alert{0.50}(84) + \alert{0.20}(92) + \alert{0.30}x}{1.00}=90
\end{equation*}
(The sum of the weights is 1.00, or 100% of Kwan’s grade.) Multiply both sides of the equation by \(1.00\) to get
\begin{equation*}
0.50(84) + 0.20(92) + 0.30x = 1.00(90)
\end{equation*}

Solve the equation. Simplify the left side first.
\begin{align*}
60.4 + 0.30x \amp = 90\amp\amp\blert{\text{Subtract 60.4 from both sides.}}\\
0.30x \amp = 29.6\amp\amp\blert{\text{Divide both sides by 0.30.}}\\
x \amp = 98.7
\end{align*}
Kwan needs a score of \(98.7\) on the final exam to earn a grade of \(90\text{.}\)
In step 2 of Example A.27, we rewrote the formula for a weighted average in a simpler form.
Weighted Average
The sum of the weighted values equals the sum of the weights times the average value. In symbols,
\begin{equation*}
w_1x_1 + w_2x_2 + \cdots + w_nx_n = W\, \overline{x}
\end{equation*}
where \(W\) is the sum of the weights.
This form is particularly useful for solving problems involving mixtures.
Example A.28
The vet advised Delbert to feed his dog Rollo with kibble that is no more than \(8\%\) fat. Rollo likes JuicyBits, which are \(15\%\) fat. LeanMeal is more expensive, but it is only \(5\%\) fat. How much LeanMeal should Delbert mix with \(50\) pounds of JuicyBits to make amixture that is \(8\%\) fat?
Solution
Let \(p\) represent the number of pounds of LeanMeal needed.

In this problem, we want the weighted average of the fat contents in the two kibbles to be \(8\%\text{.}\) The weights are the number of pounds of each kibble we use. It is often useful to summarize the given information in a table.
\(\) 
\(\%\) fat 
Total pounds 
Pounds of fat 
Juicy Bits 
\(15\%\) 
\(50\) 
\(0.15(50)\) 
LeanMeal 
\(5\%\) 
\(p\) 
\(0.05p\) 
Mixture 
\(8\%\) 
\(50+p\) 
\(0.08(50+p)\) 
The amount of fat in the mixture must come from adding the amounts of fat in the two ingredients. This gives us an equation,
\begin{equation*}
0.15(50) + 0.05p = 0.08(50 + p)
\end{equation*}
This equation is an example of the formula for weighted averages.

Simplify each side of the equation, using the distributive law on the right side, then solve.
\begin{align*}
7.5 + 0.05p \amp = 4 + 0.08p \amp\amp\blert{\text{Subtract}~4+0.05p~\text{from both sides.}}\\
3.5 \amp = 0.03p \amp\amp \blert{\text{Divid both sides by}~ 0.03.}\\
p \amp = 116.\overline{6}
\end{align*}
Delbert should mix \(116\frac{2}{3}\) pounds of LeanMeal with \(50\) pounds of JuicyBits to make a mixture that is \(8\%\) fat.