1.1: Variable Expressions
Learning Objectives
 Evaluate algebraic expressions.
 Evaluate algebraic expressions with exponents.
Introduction – The Language of Algebra
Do you like to do the same problem over and over again? No? Well, you are not alone. Algebra was invented by mathematicians so that they could solve a problem once and then use that solution to solve a group of similar problems. The big idea of algebra is that once you have solved one problem you can generalize that solution to solve other similar problems.
In this course, we'll assume that you can already do the basic operations of arithmetic. In arithmetic, only numbers and their arithmetical operations (such as
The letter
Using variables offers advantages over solving each problem “from scratch”:
 It allows the general formulation of arithmetical laws such as
a+b=b+a for all real numbersa andb .  It allows the reference to “unknown” numbers, for instance: Find a number
x such that3x+1=10 .  It allows shorthand writing about functional relationships such as, “If you sell
x tickets, then your profit will be3x−10 dollars, orf(x)=3x−10 ,” where “f ” is the profit function, andx is the input (i.e. how many tickets you sell).
Example 1
Write an algebraic expression for the perimeter and area of the rectangle as follows.
To find the perimeter, we add the lengths of all 4 sides. We can start at the topleft and work clockwise. The perimeter,
We are adding
You are probably familiar with using
It's customary in algebra to omit multiplication symbols whenever possible. For example,
Area is length multiplied by width. In algebraic terms we get the expression:
Note: An example of a variable expression is
In the above example, there is no simpler form for these equations for the perimeter and area. They are, however, perfectly general forms for the perimeter and area of a rectangle. They work whatever the numerical values of the length and width of some particular rectangle are. We would simply substitute values for the length and width of a real rectangle into our equation for perimeter and area. This is often referred to as substituting (or plugging in) values. In this chapter we will be using the process of substitution to evaluate expressions when we have numerical values for the variables involved.
Evaluate Algebraic Expressions
When we are given an algebraic expression, one of the most common things we will have to do with it is evaluate it for some given value of the variable. The following example illustrates this process.
Example 2
Let
To find the solution, substitute 12 for
The reason we place the substituted value in parentheses is twofold:
 It will make worked examples easier for you to follow.
 It avoids any confusion that would arise from dropping a multiplication sign:
2⋅12=2(12)≠212 .
Example 3
Let
Solution
Example 4
Let
Solution
Many expressions have more than one variable in them. For example, the formula for the perimeter of a rectangle in the introduction has two variables: length
Example 5
The area of a trapezoid is given by the equation
To find the solution to this problem we simply take the values given for the variables,
\begin{align*}A & = \frac{h} {2}(a + b) & & \text{Substitute} \ 10 \ \text{for} \ a, \ 15 \ \text{for} \ b \ \text{and} \ 8 \ \text{for} \ h.\\ A & = \frac{8} {2}(10 + 15) & & \text{Evaluate piece by piece}. \ (10 + 15) = 25; \frac{8} {2} = 4\\ A & = 4(25) = 100\end{align*}
Solution: The area of the trapezoid is 100 square centimeters.
Example 6
Find the value of \begin{align*} \frac{1} {9} (5x + 3y + z) \end{align*} when \begin{align*}x = 7, y = 2\end{align*} and \begin{align*}z = 11\end{align*}.
Let's plug in values for \begin{align*}x, y\end{align*} and \begin{align*}z\end{align*} and then evaluate the resulting expression.
\begin{align*}& \frac{1} {9} \left(5(7) + 3(2) + (11) \right) & & \text{Evaluate the individual terms inside the parentheses}.\\ & \frac{1} {9} \left(35 + (6) + 11 \right) & & \text{Combine terms inside parentheses}.\\ & \frac{1} {9}(40) = \frac{40} {9} \approx 4.44 \end{align*}
Solution \begin{align*} \approx 4.44\end{align*} (rounded to the nearest hundredth)
Example 7
The total resistance of two electronics components wired in parallel is given by
\begin{align*}\frac{R_1R_2} {R_1 + R_2}\end{align*}
where \begin{align*}R_{1}\end{align*} and \begin{align*}R_{2}\end{align*} are the individual resistances (in ohms) of the two components. Find the combined resistance of two such wired components if their individual resistances are 30 ohms and 15 ohms.
Solution
\begin{align*}& \frac{R_1R_2} {R_1 + R_2} & & \text{Substitute the values} \ R_1 =30 \ \text{and} \ R_2 = 15.\\ & \frac{(30)(15)} {30 + 15} = \frac{450} {45} = 10 \ \text{ohms}\end{align*}
The combined resistance is 10 ohms.
Evaluate Algebraic Expressions with Exponents
Many formulas and equations in mathematics contain exponents. Exponents are used as a shorthand notation for repeated multiplication. For example:
\begin{align*}2 \cdot 2 & = 2^{2}\\ 2 \cdot 2 \cdot 2 & = 2^{3}\end{align*}
The exponent stands for how many times the number is used as a factor (multiplied). When we deal with integers, it is usually easiest to simplify the expression. We simplify:
\begin{align*}& 2^{2}=4\\ & \ \text{and}\\ & 2^{3}=8\end{align*}
However, we need exponents when we work with variables, because it is much easier to write \begin{align*}x^{8}\end{align*} than \begin{align*}x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x\end{align*}.
To evaluate expressions with exponents, substitute the values you are given for each variable and simplify. It is especially important in this case to substitute using parentheses in order to make sure that the simplification is done correctly.
Example 8
The area of a circle is given by the formula \begin{align*}A=\pi r^{2}\end{align*}. Find the area of a circle with radius \begin{align*}r = 17 \ inches\end{align*}.
Substitute values into the equation.
\begin{align*}A & =\pi r^{2} & & \text{Substitute } 17 \ \text{for} \ r.\\ A & =\pi (17)^{2} & & \pi \cdot 17 \cdot 17 = 907.9202 \ldots \ \text{ Round to} \ 2 \ \text{decimal places}.\end{align*}
The area is approximately 907.92 square inches.
Example 9
Find the value of \begin{align*} 5x^24y\end{align*} for \begin{align*} x=4\end{align*} and \begin{align*} y=5 \end{align*}.
Substitute values in the following:
\begin{align*}5x^2 4y & = 5(4)^24(5) & & \text{Substitute} \ x=4 \ \text{and} \ y=5.\\ & =5(16)4(5) & & \text{Evaluate the exponent} \ (4)^2 = 16.\\ & = 8020\\ & = 60\end{align*}
Example 10
Find the value of \begin{align*} 2x^2 3x^2 +5\end{align*}, for \begin{align*} x=5\end{align*}.
Substitute the value of \begin{align*} x\end{align*} in the expression:
\begin{align*}2x^2 3x^2 +5 & = 2(5)^3 3(5)^2+5 && \text{Substitute} \ 5 \ \text{for} \ x.\\ & =2(125)3(25)+5 & & \text{Evaluate exponents} \ (5)^3 =(5)(5)(5)=125\\ &&& \text{and} \ (5)^2=(5)(5)=25 \\ & = 25075+5 \\ & = 320 \end{align*}
Example 11
Find the value of \begin{align*} \frac{x^2y^3} {x^3 + y^2}\end{align*}, for \begin{align*} x=2 \end{align*} and \begin{align*} y=4\end{align*}.
Substitute the values of \begin{align*} x\end{align*} and \begin{align*} y\end{align*} in the following.
\begin{align*}\frac{x^2y^3} {x^3 + y^2} & = \frac{(2)^2(4)^3} {(2)^3 + (4)^2}& & \text{Substitute} \ 2 \ \text{for} \ x \ \text{and} 4 \ \text{for} \ y.\\ \frac{4(64)} {8 + 16} & = \frac{256} {24} = \frac{32} {3}& & \text{Evaluate expressions}: (2)^2 =(2)(2)=4 \ \text{and} \ (2)^3=(2)(2)(2)=8.\\ & & & (4)^2 =(4)(4)=16 \ \text{and} \ (4)^3 = (4)(4)(4) =64.\end{align*}
Example 12
The height \begin{align*}(h)\end{align*} of a ball in flight is given by the formula: \begin{align*} h= 32t^2 + 60t+20\end{align*}, where the height is given in feet and the time \begin{align*}(t)\end{align*} is given in seconds. Find the height of the ball at time \begin{align*} t=2 \ seconds\end{align*}.
Solution
\begin{align*}h & = 32t^2 +60t+20 & & \text{Substitute} \ 2 \ \text{for} \ t.\\ & = 32(2)^2 + 60(2) + 20 \\ & = 32(4) + 60(2) + 20 \\ & = 12 \ feet\end{align*}
Review Questions
Write the following in a more condensed form by leaving out a multiplication symbol.
 \begin{align*} 2 \times 11x\end{align*}
 \begin{align*} 1.35 \cdot y\end{align*}
 \begin{align*} 3 \times \frac{1} {4}\end{align*}
 \begin{align*} \frac{1} {4} \cdot z\end{align*}
Evaluate the following expressions for \begin{align*}a=3, b=2, c=5\end{align*} and \begin{align*}d=4\end{align*}.
 \begin{align*} 2a + 3b\end{align*}
 \begin{align*} 4c + d\end{align*}
 \begin{align*} 5ac  2b\end{align*}
 \begin{align*} \frac{2a} {c  d}\end{align*}
 \begin{align*} \frac{3b} {d}\end{align*}
 \begin{align*} \frac{a  4b} {3c + 2d}\end{align*}
 \begin{align*} \frac{1} {a + b}\end{align*}
 \begin{align*} \frac{ab} {cd}\end{align*}
Evaluate the following expressions for \begin{align*}x=1, y=2, z=3,\end{align*} and \begin{align*}w=4\end{align*}.
 \begin{align*} 8x^3\end{align*}
 \begin{align*} \frac{5x^2} {6z^3}\end{align*}
 \begin{align*} 3z^2  5w^2\end{align*}
 \begin{align*} x^2  y^2\end{align*}
 \begin{align*} \frac{z^3 + w^3} {z^3  w^3}\end{align*}
 \begin{align*} 2x^2  3x^2 + 5x 4\end{align*}
 \begin{align*} 4w^3 + 3w^2  w + 2\end{align*}
 \begin{align*} 3 + \frac{1} {z^2}\end{align*}
 The weekly cost \begin{align*}C\end{align*} of manufacturing \begin{align*}x\end{align*} remote controls is given by the formula \begin{align*}C=2000+3x\end{align*}, where the cost is given in dollars.
 What is the cost of producing 1000 remote controls?
 What is the cost of producing 2000 remote controls?
 The volume of a box without a lid is given by the formula: \begin{align*}V=4x(10x)^2\end{align*} where \begin{align*}x\end{align*} is a length in inches and \begin{align*}V\end{align*} is the volume in cubic inches.
 What is the volume when \begin{align*}x=2\end{align*}?
 What is the volume when \begin{align*}x=3\end{align*}?
Review Answers
 \begin{align*} 22x\end{align*}
 \begin{align*} 1.35y\end{align*}
 \begin{align*} \frac{3} {4}\end{align*}
 \begin{align*} \frac{z} {4}\end{align*}
 0
 16
 79
 \begin{align*} \frac{2} {3}\end{align*}
 \begin{align*} \frac{3} {2}\end{align*}
 \begin{align*} \frac{11} {7}\end{align*}
 1
 \begin{align*} \frac{3} {10}\end{align*}
 8
 \begin{align*} \frac{5} {162}\end{align*}
 53
 3
 \begin{align*}\frac{37}{91}\end{align*}
 14
 302

\begin{align*} 3\frac{1}{9}\end{align*}
 $5000
 $8000
 \begin{align*}512\ in^3\end{align*}
 \begin{align*}588\ in^3\end{align*}
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