1.3 Use variables and algebraic symbols

Scott McDaniel

1.3 Use variables and algebraic symbols

Learning Objectives

Evaluate algebraic expressions for given values.
Translate word phrases and sentences into algebraic expressions.

Evaluating Algebraic Expressions

We are often presented with expressions that contain constants and variables and are then given specific values for the variable(s).

Example 1.3.1

A pizza costs $12 plus $1.50 for each topping. How much would you pay for a pizza with:

3 toppings?
5 toppings?
Can you write a formula for the cost (C) of a pizza based on the number of toppings (t)?

Solution

3 toppings → $12 + 1.5(3) = 12 + 4.50 = 16.50$
5 toppings → $12 + 1.5(5) = 12 + 7.50 = 19.50$
Formula → $C = 12 + 1.5t$

Variables and Algebraic Expressions

A variable is a symbol (often a letter such as x, y, or z) that can stand for different numbers.

An algebraic expression is a combination of numbers, variables, operation symbols, and grouping symbols (like parentheses or fraction bars).

Caution

$3m^2$ means $3 \cdot m^2$ ; it does not mean $(3m)(3m)$ .

If you want to square the entire product, use parentheses: $(3m)^2$ .

To evaluate an expression means to find its value after substituting numbers for variables. Be sure to use the order of operations (PEMDAS).

Note: At this stage, variables are just symbols that stand in for numbers. We are not solving equations yet.

Example 1.3.2

Evaluate $9x + 2$ for $x = 4$ .

Solution

$9(4) + 2 = 36 + 2 = 38$

Example 1.3.3

Evaluate $5 + 2x$ for $x = -2$ .

Solution

$5 + 2(-2) = 5 + (-4) = 1$

Example 1.3.4

Evaluate $20 - 2x^2$ for $x = -3$ .

Solution

$20 - 2(-3)^2 = 20 - 2(9) = 20 - 18 = 2$

Example 1.3.5

Evaluate $3x^2 + 2x - 4$ for $x = -1$ .

Solution

$3(-1)^2 + 2(-1) - 4 = 3(1) + (-2) - 4 = 3 + (-2) + (-4) = -3$

Example 1.3.6

Evaluate $\frac{3x - 1}{3 - 2x}$ for $x = 5$ .

Solution

$\frac{3(5) - 1}{3 - 2(5)} = \frac{15 - 1}{3 - 10} = \frac{14}{-7} = -2$

Example 1.3.7

Evaluate $5a + 4b$ for $a = 2$ and $b = 3$ .

Solution

$5(2) + 4(3) = 10 + 12 = 22$

Example 1.3.8

Evaluate $\frac{5(a + b) - 1}{c}$ for $a = 2$ , $b = 3$ , and $c = 6$ .

Solution

$\frac{5(2 + 3) - 1}{6} = \frac{5(5) - 1}{6} = \frac{25 - 1}{6} = \frac{24}{6} = 4$

Try It

Translate Sentences to Algebraic Expressions

We often need to convert words into algebraic expressions. This is the first step in solving applied problems.

Translating Word Phrases to Algebraic Expressions

Need to know

Sum → add Example: $5 + 4 = 9$
Difference → subtract Example: $5 - 4 = 1$
Product → multiply Example: $5 \times 4 = 20$
Quotient → divide Example: $\frac{5}{4} = 1.25$

Note: “Less than” reverses order.
“20 less than x” → $x - 20$

Example 1.3.9

Write each word phrase as an algebraic expression using x as the variable.

The sum of a number and 9
The product of a number and 6

Solution

$x + 9$
$6x$

Example 1.3.10

Write each word phrase as an algebraic expression using x as the variable.

The difference of $7x$ and 9
The quotient of $6x^2$ and 9

Solution

$7x - 9$
$\frac{6x^2}{9}$

Example 1.3.11

Write each word phrase as an algebraic expression using x as the variable.

20 less than 100
20 less than $10x$

Solution

$100 - 20$
$10x - 20$

Example 1.3.12

Write the algebraic expression.

The length of a rectangle is 9 less than its width.

Solution

Let $w$ represent the width.

Length = $w - 9$

Try It

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Evaluating Algebraic Expressions

Variables and Algebraic Expressions

Translate Sentences to Algebraic Expressions

Translating Word Phrases to Algebraic Expressions

Need to know

License

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