Section 6.1 The Fundamentals of Algebra Math in Our World

Learning Objectives  Identify terms and coefficients.  Simplify algebraic expressions.  Evaluate algebraic expressions.  Apply evaluating expressions to real-world situations.

Algebra Algebra is a branch of mathematics that generalizes the concepts of arithmetic by applying them to quantities that are allowed to vary. A variable, usually represented by a letter, is a quantity that can change. It represents unknown values in a situation. An algebraic expression is a combination of variables, numbers, operation symbols, and grouping symbols. Some examples of algebraic expressions are

Algebraic Expressions Algebraic expressions are made up of one or more terms. Terms are the pieces in an expression that are separated by addition or subtraction signs. In the expression 8x 2 + 6x - 3, each of 8x 2, 6x, and - 3 is a term. The expression 111 has just one term, namely 111. Every term has a numerical coefficient, or just coefficient. This is the number part of a term, like the 8 in 8x 2.

EXAMPLE 1 Identifying Terms and Coefficients Identify the terms of the algebraic expression, and the coefficient for each term. The expression has four terms: - 8y, which has coefficient - 8; 5/x 3, which has coefficient 5/2; xy, which has coefficient 1; and 3, which has coefficient 3. (When no number appears in a term, as in xy, the coefficient is 1; we just don’t need to write it.) SOLUTION

EXAMPLE 2 The Distributive Property Use the distributive property to multiply out the parentheses. (a) 5(3x + 7)(b) - 3(6A - 7B + 10)

EXAMPLE 2 The Distributive Property (a)Distributing the 5 to each term inside the parentheses, we get (b)This time we have to be careful because the term we’re distributing is negative. SOLUTION

Simplifying Algebraic Expressions When two terms have the same variables with the same exponents, we will call them like terms. To add or subtract like terms (i.e., combine like terms), add or subtract the numerical coefficients of the like terms.

EXAMPLE 3 Combining Like Terms Combine like terms for each, if possible. (a)9x – 20x (b)3x 2 + 8x 2 – 2x 2 (c)6x + 8x 2

EXAMPLE 3 Combining Like Terms (a)9x – 20x = (9 – 20)x = -11x (Found by subtracting 9 – 20) (b) 3x 2 + 8x 2 – 2x 2 = (3 + 8 – 2)x 2 = 9x 2 (Found by adding and subtracting – 2) (c) 6x + 8x 2 (These terms cannot be combined since they are not like terms.) SOLUTION

EXAMPLE 4 Combining Like Terms Simplify each expression. (a) 9x – 7y + 18 – y – 10x (b) 3x 3 + 4x 2 – 6x + 10 – 7x 2 + 4x 3 + 2x – 6

EXAMPLE 4 Combining Like Terms (a)Combine like terms: The answer is – x – y – 9. (b) Combine like terms: The answer is 7x 3 – 3x 2 – 4x + 4. SOLUTION

EXAMPLE 5 Simplifying an Algebraic Expression Simplify the expression 8(3x 2 + 5) + 3(2 – x) – (5x 2 + x). First, we multiply out the parentheses. We can think of the last set as – 1(5x 2 + x). 8(3x 2 + 5) = 24x (2 – x) = 6 – 3x – 1(5x 2 + x) = 5x 2 – x Now we combine like terms: 24x 2 + 5x 2 – 3x – x = 19x 2 – 4x + 46 SOLUTION

Evaluating Algebraic Expressions Algebraic expressions almost always contain variables, which can be any number. But when we substitute numbers in for the variables, the result is an arithmetic problem. Finding the value of this problem is called evaluating the expression.

EXAMPLE 6 Evaluating an Expression Evaluate 9x – 3 when x = 5. Substitute 5 for the variable x, then perform the calculation: 9x – 3 = 9(5) – 3 Multiply first. = 45 – 3 Subtract. = 42 The value of 9x – 3 when x = 5 is 42. SOLUTION

EXAMPLE 7 Evaluating an Expression with Two Variables Evaluate 5x 2 – 7 y + 2 when x = – 3 and y = 6. Substitute – 3 for x and 6 for y into the expression, then simplify: 5x 2 – 7y + 2 = 5(– 3) 2 – 7(6) + 2 Exponents first. = 5(9) – 7(6) + 2 Multiply twice. = 45 – Subtract. = 5 The value of 5x 2 – 7y + 2 when x = – 3 and y = 6 is 5. SOLUTION

EXAMPLE 8 Finding Commission on Sales A salesperson at a popular clothing store gets a $600 monthly salary and a 10% commission on everything she sells. The expression 0.10x describes the amount of money she earns each month, where x represents the dollar amount of sales. If she had net sales of $13,240 in July, how much did she earn?

Since we were told that x represents monthly sales, and that her sales for July were $13,240, we substitute 13,240 in for x, then simplify. 0.10(13,240) = 1, = 1,924 The salesperson earned $1,924 in July. EXAMPLE 8 Finding Commission on Sales SOLUTION

EXAMPLE 9 Computing Total Cost Including Tax The state of Florida has a sales tax of 6%. (a) Write an expression for the total cost of an item purchased in Florida, including sales tax. The variable should represent the cost of the item before tax. (b) John bought an iPhone at a store in Hollywood, Florida. The price was $349. What was the total cost John paid, including tax?

(a) We’ll use the variable x to represent the cost of the item before tax. The sales tax will be 6% of x, or 0.06x. That makes the total cost x x, which simplifies to 1.06x. (b) Substitute 349 in for x: 1.06(349) = $ EXAMPLE 9 Computing Total Cost Including Tax SOLUTION

EXAMPLE 10 Computing a Discount Price While shopping at her favorite department store, Carmen found a dress she’s been hoping to buy on the 40%-off clearance rack. (a) Write an algebraic expression representing the new price of the dress before tax, then one for the total price including tax. Use 6% as the sales tax rate. (b) If the original price of the dress was $59, find the discounted price, and the total amount that Sally would pay including tax.

(a) Let x represent the original price of the dress. Then the amount of the discount is 40% of x, or 0.4x. So the sale price is x – 0.4x, which simplifies to 0.6x. With a sales tax rate of 6%, the total cost will be 0.6x (0.6x) = 0.6x x = 0.636x (b) In our expressions in part (a), the variable x represented the original price of the dress, so we substitute 59 in for x. The discount price is 0.6(59) = $35.40, and The total cost including tax is 0.636(59) = $ EXAMPLE 10 Computing a Discount Price SOLUTION Discount Price 6% Sales Tax

EXAMPLE 11 Using the Formula for Distance The distance in miles an automobile travels is given by the formula d = rt, where r is the rate (or speed) in miles per hour and t is the time in hours. How far will an automobile travel in 6 hours at a rate of 55 miles per hour? In the formula d = rt, substitute 55 for r and 6 for t and evaluate. d = rt d = 55(6) d = 330 The automobile will travel 330 miles in 6 hours. SOLUTION