1.1: Variable Expressions

Difficulty Level: At Grade Created by: CK-12

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 \begin{align*}+, -, \times, \div\end{align*}) occur. In algebra, numbers (and sometimes processes) are denoted by symbols (such as \begin{align*}x, y, a, b, c, \ldots\end{align*}). These symbols are called variables.

The letter \begin{align*}x\end{align*}, for example, will often be used to represent some number. The value of \begin{align*}x\end{align*}, however, is not fixed from problem to problem. The letter \begin{align*}x\end{align*} will be used to represent a number which may be unknown (and for which we may have to solve) or it may even represent a quantity which varies within that problem.

Using variables offers advantages over solving each problem “from scratch”:

• It allows the general formulation of arithmetical laws such as \begin{align*}a+b=b+a\end{align*} for all real numbers \begin{align*}a\end{align*} and \begin{align*}b\end{align*}.
• It allows the reference to “unknown” numbers, for instance: Find a number \begin{align*}x\end{align*} such that \begin{align*} 3x+1=10 \end{align*}.
• It allows short-hand writing about functional relationships such as, “If you sell \begin{align*}x\end{align*} tickets, then your profit will be \begin{align*}3x-10\end{align*} dollars, or \begin{align*}f(x)=3x-10\end{align*},” where “ \begin{align*}f\end{align*} ” is the profit function, and \begin{align*}x\end{align*} 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 top-left and work clockwise. The perimeter, \begin{align*}P\end{align*}, is therefore:

\begin{align*}P=l+w+l+w\end{align*}

We are adding \begin{align*}2 l\end{align*}'s and \begin{align*}2 w\end{align*}'s. Would say that:

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

You are probably familiar with using \begin{align*}\cdot\end{align*} instead of \begin{align*}\times\end{align*} for multiplication, so you may prefer to write:

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

It's customary in algebra to omit multiplication symbols whenever possible. For example, \begin{align*} 11x \end{align*} means the same thing as \begin{align*} 11 \cdot x \end{align*} or \begin{align*} 11 \times x \end{align*}. We can therefore write the expression for \begin{align*}P\end{align*} as:

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

Area is length multiplied by width. In algebraic terms we get the expression:

\begin{align*}A = l \times w \qquad \rightarrow \qquad A = l \cdot w \qquad \rightarrow \qquad A = lw\end{align*}

Note: An example of a variable expression is \begin{align*}2l+2w\end{align*}; an example of an equation is \begin{align*}P=2l+2w\end{align*}. The main difference between equations and expressions is the presence of an equals sign (=).

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 \begin{align*}x = 12\end{align*}. Find the value of \begin{align*}2x-7\end{align*}.

To find the solution, substitute 12 for \begin{align*} x \end{align*} in the given expression. Every time we see \begin{align*}x\end{align*} we will replace it with 12. Note: At this stage we place the value in parentheses:

\begin{align*}2x-7 & = 2(12)-7\\ & = 24-7\\ & = 17\end{align*}

The reason we place the substituted value in parentheses is twofold:

1. It will make worked examples easier for you to follow.
2. It avoids any confusion that would arise from dropping a multiplication sign: \begin{align*} 2 \cdot 12 = 2(12) \neq 212 \end{align*}.

Example 3

Let \begin{align*}x = -1\end{align*}. Find the value of \begin{align*} -9x + 2\end{align*}.

Solution

\begin{align*}-9(-1)+2 & = 9 + 2 \\ & = 11\end{align*}

Example 4

Let \begin{align*}y = -2\end{align*}. Find the value of \begin{align*} \frac{7} {y} - 11y + 2\end{align*}.

Solution

\begin{align*}\frac{7} {(-2)} - 11(-2) + 2 &= -3\frac{1} {2} + 22 + 2 \\ &= 24 - 3 \frac{1} {2} \\ &= 20 \frac{1} {2} \end{align*}

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 \begin{align*}(l)\end{align*} and width \begin{align*}(w)\end{align*}. In these cases be careful to substitute the appropriate value in the appropriate place.

Example 5

The area of a trapezoid is given by the equation \begin{align*}A = \frac{h}{2} (a + b)\end{align*}. Find the area of a trapezoid with bases \begin{align*} a=10 \ cm, b=15 \ cm\end{align*} and height \begin{align*}h=8 \ cm\end{align*}.

To find the solution to this problem we simply take the values given for the variables, \begin{align*} a, b \end{align*} and \begin{align*} h \end{align*}, and plug them in to the expression for \begin{align*} A \end{align*}:

\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 short-hand 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^2-4y\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)^2-4(5) & & \text{Substitute} \ x=-4 \ \text{and} \ y=5.\\ & =5(16)-4(5) & & \text{Evaluate the exponent} \ (-4)^2 = 16.\\ & = 80-20\\ & = 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 \\ & = -250-75+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.

1. \begin{align*} 2 \times 11x\end{align*}
2. \begin{align*} 1.35 \cdot y\end{align*}
3. \begin{align*} 3 \times \frac{1} {4}\end{align*}
4. \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*}.

1. \begin{align*} 2a + 3b\end{align*}
2. \begin{align*} 4c + d\end{align*}
3. \begin{align*} 5ac - 2b\end{align*}
4. \begin{align*} \frac{2a} {c - d}\end{align*}
5. \begin{align*} \frac{3b} {d}\end{align*}
6. \begin{align*} \frac{a - 4b} {3c + 2d}\end{align*}
7. \begin{align*} \frac{1} {a + b}\end{align*}
8. \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*}.

1. \begin{align*} 8x^3\end{align*}
2. \begin{align*} \frac{5x^2} {6z^3}\end{align*}
3. \begin{align*} 3z^2 - 5w^2\end{align*}
4. \begin{align*} x^2 - y^2\end{align*}
5. \begin{align*} \frac{z^3 + w^3} {z^3 - w^3}\end{align*}
6. \begin{align*} 2x^2 - 3x^2 + 5x -4\end{align*}
7. \begin{align*} 4w^3 + 3w^2 - w + 2\end{align*}
8. \begin{align*} 3 + \frac{1} {z^2}\end{align*}
9. 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.
1. What is the cost of producing 1000 remote controls?
2. What is the cost of producing 2000 remote controls?
10. The volume of a box without a lid is given by the formula: \begin{align*}V=4x(10-x)^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.
1. What is the volume when \begin{align*}x=2\end{align*}?
2. What is the volume when \begin{align*}x=3\end{align*}?

1. \begin{align*} 22x\end{align*}
2. \begin{align*} 1.35y\end{align*}
3. \begin{align*} \frac{3} {4}\end{align*}
4. \begin{align*} \frac{z} {4}\end{align*}
5. 0
6. 16
7. -79
8. \begin{align*} \frac{-2} {3}\end{align*}
9. \begin{align*} \frac{-3} {2}\end{align*}
10. \begin{align*} \frac{-11} {7}\end{align*}
11. -1
12. \begin{align*} \frac{3} {10}\end{align*}
13. -8
14. \begin{align*} \frac{-5} {162}\end{align*}
15. -53
16. -3
17. \begin{align*}\frac{37}{-91}\end{align*}
18. -14
19. 302
20. \begin{align*} 3\frac{1}{9}\end{align*}
1. $5000 2.$8000
1. \begin{align*}512\ in^3\end{align*}
2. \begin{align*}588\ in^3\end{align*}

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