# 1.1: Variable Expressions

**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 **variables** (which are usually letters, such as

For example, we might use the letter

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

- Formulate arithmetical laws such as
a+b=b+a for all real numbersa andb . - Refer to “unknown” numbers. For instance: find a number
x such that3x+1=10 . - Write more compactly about functional relationships such as, “If you sell
x tickets, then your profit will be3x−10 dollars, or “f(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 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

We are adding

It's customary in algebra to omit multiplication symbols whenever possible. For example,

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

Note: **variable expression**; **equation**. The main difference between expressions and **equa**tions is the presence of an **equa**ls 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. Find the value of 2x−7.*

To find the solution, we substitute 12 for

**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

**Example 3**

*Let y=−2. Find the value of 7y−11y+2.*

**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 A=h2(a+b). Find the area of a trapezoid with bases a=10 cm and b=15 cm and height h=8 cm.*

To find the solution to this problem, we simply take the values given for the variables

**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:

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:

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

- 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^3 - 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?
- What is the cost of producing 2500 remote controls?

- 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.
- What is the volume when \begin{align*}x = 2\end{align*}?
- What is the volume when \begin{align*}x = 3\end{align*}?

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