Chapter 6 Section 1 - Slide 1 Copyright © 2009 Pearson Education, Inc. Chapter 6 Section 1 - Slide 1 1. Algebra 2. Functions

Chapter 6 Section 1 - Slide 2 Copyright © 2009 Pearson Education, Inc. Chapter 6 Section 1 - Slide 2 Order of Operations

Chapter 6 Section 1 - Slide 3 Copyright © 2009 Pearson Education, Inc. Definitions Algebra: a generalized form of arithmetic. Variables: letters used to represent numbers that vary Constant: symbol that represents a specific number Algebraic expression: a phrase including variables, numbers, parentheses, and operations. Examples:

Chapter 6 Section 1 - Slide 4 Copyright © 2009 Pearson Education, Inc. Order of Operations P.E.M.D.A.S. 1. Parentheses 2. Exponents 3. Multiplication & Division (left to right) 4. Addition & Subtraction (left to right)

Chapter 6 Section 1 - Slide 5 Copyright © 2009 Pearson Education, Inc. Example: Evaluating an Expression Evaluate the expression x 2 + 4x + 5 for x = 3. Solution: x 2 + 4x + 5 = (3) + 5 = 9 + 4(3) + 5 Exponents = Multiplication = 26 Addition

Chapter 6 Section 1 - Slide 6 Copyright © 2009 Pearson Education, Inc. Example: Substituting for Two Variables Evaluate when x = 3 and y = 4. Solution:

Chapter 6 Section 1 - Slide 7 Copyright © 2009 Pearson Education, Inc. Chapter 6 Section 1 - Slide 7 Linear Equations in One Variable

Chapter 6 Section 1 - Slide 8 Copyright © 2009 Pearson Education, Inc. Definitions Terms are parts that are added in an algebraic expression. Coefficient is the numerical part of a term. Like terms are terms that have the same variables with the same exponents on the variables. Unlike terms have different variables or different exponents on the variables.

Chapter 6 Section 1 - Slide 9 Copyright © 2009 Pearson Education, Inc. Properties of the Real Numbers Associative property of multiplication (ab)c = a(bc) Associative property of addition (a + b) + c = a + (b + c) Commutative property of multiplication ab = ba Commutative property of addition a + b = b + a Distributive propertya(b + c) = ab + ac

Chapter 6 Section 1 - Slide 10 Copyright © 2009 Pearson Education, Inc. Example: Combine Like Terms 1. 8x + 4x = (8 + 4)x = 12x 2. 3x y  4 + 7x = 3x + 7x + 6y = (3 + 7)x + 6y + (2  4) = ???

Chapter 6 Section 1 - Slide 11 Copyright © 2009 Pearson Education, Inc. Solving Equations Addition Property of Equality If a = b, then a + c = b + c for all real numbers a, b, and c. Find the solution to the equation x  9 = 24 x  = x = 33 Check: x  9 =  9 = 24 ? 24 = 24 true

Chapter 6 Section 1 - Slide 12 Copyright © 2009 Pearson Education, Inc. Solving Equations (continued) Subtraction Property of Equality If a = b, then a  c = b  c for all real numbers a, b, and c. Find the solution to the equation x + 12 = 31 x + 12  12 = 31  12 x = 19 Check: x + 12 = = 31 ? 31 = 31 true

Chapter 6 Section 1 - Slide 13 Copyright © 2009 Pearson Education, Inc. Solving Equations (continued) Multiplication Property of Equality If a = b, then a c = b c for all real numbers a, b, and c, where c  0. Find the solution to the equation

Chapter 6 Section 1 - Slide 14 Copyright © 2009 Pearson Education, Inc. Solving Equations (continued) Division Property of Equality If a = b, then for all real numbers a, b, and c, c  0. Find the solution to the equation 4x = 48.

Chapter 6 Section 1 - Slide 15 Copyright © 2009 Pearson Education, Inc. General Procedure for Solving Linear Equations If the equation contains fractions, multiply both sides of the equation by the lowest common denominator. Use the distributive property to remove parentheses when necessary. Combine like terms on the same side of the equal sign when possible. Use the addition or subtraction property to collect all terms with a variable on 1 side of the equal sign and all constants on the other side of the equal sign. This process will eventually result in an equation of the form ax = b, where a and b are real numbers. Solve for the variable using the division or multiplication property. The result will be an answer in the form x = c, where c is a real number.

Chapter 6 Section 1 - Slide 16 Copyright © 2009 Pearson Education, Inc. Example: Solving Equations Solve 3x  4 = 17. X = ??? Survey Says!?

Chapter 6 Section 1 - Slide 17 Copyright © 2009 Pearson Education, Inc. Example: Solving Equations Solve 21 = 6 + 3(x + 2). X = ??? Survey Says!?

Chapter 6 Section 1 - Slide 18 Copyright © 2009 Pearson Education, Inc. Example: Solving Equations Solve 8x + 3 = 6x X = ??? Survey Says!?

Chapter 6 Section 1 - Slide 19 Copyright © 2009 Pearson Education, Inc. Example: Solving Equations Solve This statement -9=-10 is false, so the equation has no solution. The equation is called inconsistent.

Chapter 6 Section 1 - Slide 20 Copyright © 2009 Pearson Education, Inc. Example: Solving Equations Solve The statement -8 = -8 is true, so the solution is all real numbers. In other words, you can plug-in any value for x and the statement holds.

Chapter 6 Section 1 - Slide 21 Copyright © 2009 Pearson Education, Inc. Proportions A proportion is a statement of equality between two ratios. Cross Multiplication If then ad = bc, b  0, d  0.

Chapter 6 Section 1 - Slide 22 Copyright © 2009 Pearson Education, Inc. Application Example using Proportions A 50 pound bag of fertilizer will cover an area of 15,000 ft 2. How many pounds are needed to cover an area of 226,000 ft 2 ? X = ???

Chapter 6 Section 1 - Slide 23 Copyright © 2009 Pearson Education, Inc. Chapter 6 Section 1 - Slide 23 Formulas

Chapter 6 Section 1 - Slide 24 Copyright © 2009 Pearson Education, Inc. Definitions A formula is an equation that typically has a real-life application. To evaluate a formula, substitute the given values for their respective variables and then evaluate using the order of operations.

Chapter 6 Section 1 - Slide 25 Copyright © 2009 Pearson Education, Inc. Perimeter The formula for the perimeter of a rectangle is Perimeter = 2 length + 2 width or P = 2l + 2w. Use the formula to find the perimeter of a yard when l = 150 feet and w = 100 feet. P = 2l + 2w P = 2(150) + 2(100) P = P = 500 feet

Chapter 6 Section 1 - Slide 26 Copyright © 2009 Pearson Education, Inc. Example The formula for the volume of a cylinder is V =  r 2 h where  is an irrational number equal to approximately Use the formula to find the height of a cylinder with a radius of 6 inches and a volume of in 3. Let p represent pi. The height of the cylinder is approximately 5 inches.

Chapter 6 Section 1 - Slide 27 Copyright © 2009 Pearson Education, Inc. Chapter 6 Section 1 - Slide 27 Applications of Linear Equations in One Variable

Chapter 6 Section 1 - Slide 28 Copyright © 2009 Pearson Education, Inc. To Solve a Word Problem Read the problem carefully at least twice to be sure that you understand it. If possible, draw a sketch to help visualize the problem. Determine which quantity you are being asked to find. Choose a letter to represent this unknown quantity. Write down exactly what this letter represents. Write the word problem as an equation. Solve the equation for the unknown quantity. Answer the question or questions asked. Check the solution.

Chapter 6 Section 1 - Slide 29 Copyright © 2009 Pearson Education, Inc. Example The bill (parts and labor) for the repairs of a car was $ The cost of the parts was $339. The cost of the labor was $45 per hour. How many hours were billed? Let h = the number of hours billed Cost of parts + labor = total amount h = Survey Says!?

Chapter 6 Section 1 - Slide 30 Copyright © 2009 Pearson Education, Inc. Example (continued) h = ???

Chapter 6 Section 1 - Slide 31 Copyright © 2009 Pearson Education, Inc. Example Sandra Cone wants to fence in a rectangular region in her backyard for her lambs. She only has 184 feet of fencing to use for the perimeter of the region. What should the dimensions of the region be if she wants the length to be 8 feet greater than the width?

Chapter 6 Section 1 - Slide 32 Copyright © 2009 Pearson Education, Inc. Example (continued) 184 feet of fencing, length 8 feet longer than width Let x = width of region Then x + 8 = length P = 2l + 2w x + 8 x X = ??? Survey Says!?