<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 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

No one likes doing the same problem over and over again—that’s why mathematicians invented algebra. Algebra takes the basic principles of math and makes them more general, so we can solve a problem once and then use that solution to solve a group of similar problems.

In arithmetic, you’ve dealt with numbers and their arithmetical operations (such as +, , ×, ÷\begin{align*}+, \ -, \ \times, \ \div\end{align*}). In algebra, we use symbols called variables (which are usually letters, such as x, y, a, b, c, \begin{align*}x, \ y, \ a, \ b, \ c, \ \ldots\end{align*}) to represent numbers and sometimes processes.

For example, we might use the letter x\begin{align*}x\end{align*} to represent some number we don’t know yet, which we might need to figure out in the course of a problem. Or we might use two letters, like x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}, to show a relationship between two numbers without needing to know what the actual numbers are. The same letters can represent a wide range of possible numbers, and the same letter may represent completely different numbers when used in two different problems.

Using variables offers advantages over solving each problem “from scratch.” With variables, we can:

• Formulate arithmetical laws such as a+b=b+a\begin{align*}a + b = b + a \end{align*} for all real numbers a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}.
• Refer to “unknown” numbers. For instance: find a number x\begin{align*}x\end{align*} such that 3x+1=10\begin{align*}3x + 1 = 10\end{align*}.
• Write more compactly about functional relationships such as, “If you sell x\begin{align*}x\end{align*} tickets, then your profit will be 3x10\begin{align*}3x - 10\end{align*} dollars, or “f(x)=3x10\begin{align*}f(x) = 3x - 10\end{align*},” where “f\begin{align*}f\end{align*}” is the profit function, and x\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 below.

To find the perimeter, we add the lengths of all 4 sides. We can still do this even if we don’t know the side lengths in numbers, because we can use variables like l\begin{align*}l\end{align*} and w\begin{align*}w\end{align*} to represent the unknown length and width. If we start at the top left and work clockwise, and if we use the letter P\begin{align*}P\end{align*} to represent the perimeter, then we can say:

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

We are adding 2 l\begin{align*}2 \ l\end{align*}’s and 2 w\begin{align*}2 \ w\end{align*}'s, so we can say that:

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

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

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

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

A=l×wA=lwA=lw\begin{align*} A = l \times w \to A = l \cdot w \to A = lw\end{align*}

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

In the above example, we found the simplest possible ways to express the perimeter and area of a rectangle when we don’t yet know what its length and width actually are. Now, when we encounter a rectangle whose dimensions we do know, we can simply substitute (or plug in) those values in the above equations. In this chapter, we will encounter many expressions that we can evaluate by plugging in values for the variables involved.

## Evaluate Algebraic Expressions

When we are given an algebraic expression, one of the most common things we might have to do with it is evaluate it for some given value of the variable. The following example illustrates this process.

Example 2

Let x=12\begin{align*}x = 12\end{align*}. Find the value of 2x7\begin{align*}2x - 7\end{align*}.

To find the solution, we substitute 12 for x\begin{align*}x\end{align*} in the given expression. Every time we see x\begin{align*}x\end{align*}, we replace it with 12.

2x7=2(12)7=247=17\begin{align*}2x - 7 &= 2(12) - 7\\ &= 24 - 7\\ &= 17\end{align*}

Note: At this stage of the problem, we place the substituted value in parentheses. We do this to make the written-out problem easier to follow, and to avoid mistakes. (If we didn’t use parentheses and also forgot to add a multiplication sign, we would end up turning 2x\begin{align*}2x\end{align*} into 212 instead of 2 times 12!)

Example 3

Let y=2.\begin{align*}y = -2. \end{align*} Find the value of 7y11y+2\begin{align*} \frac {7} {y} - 11 y + 2 \end{align*}.

Solution

7(2)11(2)+2=312+22+2=24312=2012\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 (l)\begin{align*}(l)\end{align*} and width (w)\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 A=h2(a+b)\begin{align*} A = \frac{ h } { 2 } (a + b)\end{align*}. Find the area of a trapezoid with bases a=10 cm\begin{align*}a = 10 \ cm\end{align*} and b=15 cm\begin{align*}b = 15 \ cm\end{align*} and height h=8 cm\begin{align*}h = 8 \ cm\end{align*}.

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

A=h2(a+b)Substitute 10 for a, 15 for b, and 8 for h.A=82(10+15)Evaluate piece by piece. 10+15=25; 82=4.A=4(25)=100\begin{align*}& A = \frac { h } { 2 }(a + b) \qquad \text{Substitute} \ 10 \ \text{for} \ a, \ 15 \ \text{for} \ b, \ \text{and} \ 8 \ \text{for} \ h.\\ & A = \frac { 8 } { 2 }(10 + 15) \quad \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.

## 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:

22222=22=23\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:

2223=4=8\begin{align*} 2^2 &= 4\\ 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.

For a more detailed review of exponents and their properties, check out the video at http://www.mathvids.com/lesson/mathhelp/863-exponents---basics.

Example 5

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 \qquad \ \text{Substitute} \ 17 \ \text{for} \ r.\\ & A = \pi (17)^2 \quad \pi \cdot 17 \cdot 17 \approx 907.9202 \ldots \ \text{Round to} \ 2 \ \text{decimal places.}\end{align*}

The area is approximately 907.92 square inches.

Example 6

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 } \qquad \ \text{Substitute} \ 2 \ \text{for} \ x \ \text{and} \ -4 \ \text{for} \ y.\\ & \frac { 4(-64) } { 8 + 16 } = \frac { - 256 } { 24 } = \frac{-32}{3} \qquad \text{Evaluate expressions:} \ (2)^2 = (2)(2) = 4 \ \text{and}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \ (2)^3 = (2)(2)(2) = 8. \ (-4)^2 = (-4)(-4) = 16 \ \text{and}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \ (-4)^3 = (-4)(-4)(-4) = -64.\end{align*}

Example 7

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\\ &= -32(2)^2 + 60(2) + 20 \qquad \text{Substitute} \ 2 \ \text{for} \ t.\\ &= -32(4) + 60(2) + 20\\ &= 12\end{align*}

The height of the ball is 12 feet.

## Review Questions

1. 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*}
2. 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*}
3. 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^3 - 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*}
4. 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?
3. What is the cost of producing 2500 remote controls?
5. 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*}?

### My Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: