Section2.1Variables and Evaluating Expressions
¶A key sign that you are moving on from arithmetic to algebra is if you are using variables, discussed in this section. Any combination of numbers and variables using mathematical operations is called a mathematical expression. Some expressions are simple, and some are complicated. Some expressions are abstract, whereas some have context and meaning. One example of a simple expression with context is \(220 - a\text{,}\) which has one variable, \(a\text{,}\) and is the expression for the maximum heart rate of a person who is \(a\) years old.
Along with variables, and expressions, mathematical equations and inequalities are very important in algebra. In this section, we'll focus on expressions, and the remainder of this chapter is devoted to equations and inequalities.
Subsection2.1.1Introduction to Variables
When we want to represent an unknown or changing numerical quantity, we use a variable to do so. For example, if you'd like to analyze the gas mileage of various cars, you could let the symbol “\(g\)” represent a car's gas mileage. The mileage might be 25 mpg, 30 mpg, or some other quantity. If we agree to use mpg for the units of measure, \(g\) might be a place holder for \(25\text{,}\) \(30\text{,}\) or some other number. Since we are using a variable and not a specific number, we can analyze gas mileage for Honda Civics at the same time we analyze gas mileage for Ford Explorers.
When using variables to stand for actual physical quantities, it's good practice to use letters that clearly correspond to the quantity they represent. For example, it's wise to use \(g\) to represent gas mileage. This helps the people who might read your work in the future to understand it better.
At the same time you identify what variable you would like to use, it is very important to identify what units of measurement will go along with that variable, and clearly tell your reader this. For example, suppose you are working with \(g=25\text{.}\) A car whose gas mileage is 25 mpg is very different from a car whose gas mileage is 25 kpg (kilometer per gallon). So it would be important to state that \(g\) represents gas mileage in miles per gallon.
Exercise2.1.3
Remark2.1.4
Note that unless an algebra problem specifies which letter(s) to use, we have a choice as to which letter we choose to represent our variable(s). However without any context to a problem, \(x\text{,}\) \(y\text{,}\) and \(z\) are the most common letters used as variables, and you may see these variables (especially \(x\)) a lot.
Also note that the units we use are often determined indirectly by other quantitative information given in an algebra problem. For example, if we're told that a car has used so many gallons of gas after traveling so many miles, then it suggests we should measure gas mileage in mpg.
Subsection2.1.2Mathematical Expressions
A mathematical expression is any combination of variables and numbers using arithmetic operations. The following are all examples of mathematical expressions:
\begin{equation*} x+1\qquad 2\ell+2w\qquad\frac{\sqrt{x}}{y+1}\qquad nRT \end{equation*}Note that this definition of “mathematical expression” does not include anything with signs like these in them: \(=\text{,}\) \(\lt\text{,}\) \(\leq\text{,}\) etc.
Example2.1.5
The expression:
\begin{equation*} \frac{5}{9}(F - 32) \end{equation*}can be used to convert from degrees Fahrenheit to degrees Celsius. To do this, we need a Fahrenheit temperature, \(F\text{.}\) Then we can evaluate the expression. This means replacing its variable(s) with specific numbers and calculating the result. In this case, we can replace \(F\) with a specific number.
Let's convert the temperature 89 °F to the Celsius scale by evaluating the expression.
\begin{align*} \frac{5}{9}(F - 32) \amp= \frac{5}{9}(\substitute{89} - 32)\\ \amp= \frac{5}{9}(57)\\ \amp= \frac{285}{9}\\ \amp\approx 31.67 \end{align*}This shows us that 89 °F is equivalent to approximately 31.67 °C.
Warning2.1.6Vocabulary Usage
The steps in Example 2.1.5 are not considered to be “solving an equation,” which is language you might be tempted to use. Instead, this process was “evaluating an expression.” There is a special algebra meaning for words like “solve” and “solution” that will come soon.
In Example 2.1.5, we did use equals signs, but that was so we could assert that one expression equals another expression, which equals another expression, and so on. In the end, this example showed us that \(\frac{5}{9}(F - 32)\) evaluates to \(\approx31.67\) when \(F\) is \(89\text{.}\) It did not “solve” anything in a technical mathematical sense.
Exercise2.1.7
Try evaluating the temperature expression for yourself.
Example2.1.8Target heart rate
According to the American Heart Association, maximum heart rate in beats per minute (bpm) is given by \(220 - a\text{,}\) where \(a\) is age in years.
Determine the maximum heart rate for someone who is \(31\) years old.
Target heart rate for moderate exercise is \(50\%\) to \(70\%\) of maximum heart rate. If we want to reach \(65\%\) of an individual's maximum heart rate during moderate exercise, we'd use the expression \(0.65(220-a)\text{,}\) where \(a\) is their age in years. Determine the target heart rate at this \(65\%\) level for someone who is \(31\) years old.
Solution
Both of these parts ask us to evaluate an expression.
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Since \(a\) is defined to be age in years, we will evaluate this expression by substituting \(a\) with \(31\text{:}\)
\begin{align*} 220-a \amp= 220-\substitute{31}\\ \amp= 189 \end{align*}This tells us that the maximum heart rate for someone who is \(31\) years old is \(189\) bpm.
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We'll again substitute \(a\) with \(31\text{,}\) but this time using the target heart rate expression:
\begin{align*} 0.65(220-a) \amp= 0.65(220-\substitute{31})\\ \amp= 0.65(189)\\ \amp=122.85 \end{align*}This tells us that the target heart rate for someone who is \(31\) years old undertaking moderate exercise is \(122.85\) bpm.
Exercise2.1.9
Example2.1.10Geometry
An expression for the area of a trapezoid (shown in Figure 2.1.11) is \(\frac{1}{2}(a+b)h\text{,}\) where \(a\) and \(b\) are the parallel side lengths and \(h\) is the height. This expression is valid if \(a\text{,}\) \(b\text{,}\) and \(h\) are all measured in centimeters, inches, or any other unit for length. The result will be measured in square units, such as cm2 or in2.
Determine the area of a trapezoid for which \(a=3\,\text{cm}\text{,}\) \(b=5\,\text{cm}\text{,}\) and \(h=2\,\text{cm}\text{.}\)
Solution
Replacing each variable with the given value:
\begin{align*} \frac{1}{2}(a+b)h \amp= \frac{1}{2}(\substitute{3\text{ cm}}+\substitute{5\text{ cm}})(\substitute{2\text{ cm}})\\ \amp= \frac{1}{2}(8\text{ cm})(2\text{ cm})\\ \amp=8\text{ cm}^2 \end{align*}The area of this trapezoid is 8 cm2.
Exercise2.1.12Rising Rents
Terms and Factors
Some expressions consist of terms — parts that are added together. For example, the expression \(2x^2+3x-4\) has three terms: \(2x^2\text{,}\) \(3x\) and \(-4\text{.}\)
Some expressions consist of factors — parts that are multiplied together. For example, the term \(2x^2\) has two factors: \(2\) and \(x^2\) (with the multiplication symbol implied between them). The term \(3x\) has two factors: \(3\) and \(x\text{.}\) The expression \(x(x+1)(x-2)\) has three factors: \(x\text{,}\) \((x+1)\text{,}\) and \((x-2)\text{.}\)
Subsection2.1.3Evaluating Expressions with Exponents, Absolute Value, and Radicals
Mathematical expressions will often have exponents, absolute value bars, and radicals. This does not change the basic approach to evaluating them.
Example2.1.13Cylinder Volume
The volume \(V\text{,}\) in cm3, of a cylinder with height \(h\) and radius \(r\text{,}\) both in cm, is given by \(V=\pi r^2h\text{.}\) Assume that a can of tuna has a radius of 4 cm and a height of 2 cm.
Calculate the volume of this can in terms of \(\pi\text{.}\)
Calculate the volume of this can and round the answer to four significant digits. Use the \(\pi\) button on your calculator, or use \(\pi\approx3.1415927\text{.}\)
Solution
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Substitute \(r=4\) and \(h=2\) into the expression \(\pi r^2h\text{,}\) we have:
\begin{align*} V=\pi r^2h\amp=\pi(\substitute{4\text{ cm}})^2(\substitute{2\text{ cm}})\\ \amp=\pi (16\text{ cm}^2)(2\text{ cm})\\ \amp=\pi(32\text{ cm}^3)\\ \amp=32\pi\text{ cm}^3 \end{align*}The volume of this can is 32\(\pi\) cm3.
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For this part, we substitute \(\pi\approx3.1415927\) into out first result, \(32\pi\text{,}\) and we have:
\begin{align*} V=32\pi\amp\approx 32(\substitute{3.1415927})\\ \amp\approx 100.5309\ldots\\ \amp\approx\overbrace{100.5}^{\text{four}}309\ldots \end{align*}The volume of this can is about 100.5 cm3.
Example2.1.14Tsunami Speed
The speed of a tsunami (in meters per second) can be modeled by \(\sqrt{9.8d}\text{,}\) where \(d\) is the depth of the tsunami (in meters). Determine the speed of a tsunami that has a depth of 30 m to four significant digits.
Solution
Using \(d=30\text{,}\) we find:
\begin{align*} \sqrt{9.8d} \amp= \sqrt{9.8(\substitute{30})}\\ \amp=\sqrt{294}\\ \amp\approx \overbrace{17.14}^{\text{four}}6428\ldots \end{align*}The speed of tsunami with a depth of 30 m is about 17.15 m⁄s.
Exercise2.1.15Tent Height
Exercise2.1.16Mortgage Payments
Subsection2.1.4Evaluating Expressions with Negative Numbers
When we substitute negative numbers into an expression, it's important to use parentheses around them. Let's look at some examples.
Example2.1.17
Evaluate \(x^2\) if \(x=-2\text{.}\)
We substitute:
\begin{align*} x^2\amp=(\substitute{-2})^2\\ \amp=4 \end{align*}If we don't use parentheses, we would have:
\begin{align*} x^2\amp=-2^2\amp\text{incorrect!}\\ \amp=-4 \end{align*}The original expression takes \(x\) and squares it. With \(-2^2=-4\text{,}\) the number \(-2\) is not being squared. Since the exponent has higher priority than the negation, it's just the number \(2\) that is being squared. With \((-2)^2=4\) the number \(-2\) is being squared, which is what we would want given the expression \(x^2\text{.}\)
So it is wise to always use some parentheses when substituting in any negative number.
Exercise2.1.18
Subsection2.1.5Exercises
Evaluating Expressions
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Evaluating Expressions in Context