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Slide 6 - 1 Copyright © 2009 Pearson Education, Inc. MM150 Unit 3 Seminar Agenda Order of Operations Linear Equations Formulas Applications of Linear Equations.

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Presentation on theme: "Slide 6 - 1 Copyright © 2009 Pearson Education, Inc. MM150 Unit 3 Seminar Agenda Order of Operations Linear Equations Formulas Applications of Linear Equations."— Presentation transcript:

1 Slide 6 - 1 Copyright © 2009 Pearson Education, Inc. MM150 Unit 3 Seminar Agenda Order of Operations Linear Equations Formulas Applications of Linear Equations

2 Slide 6 - 2 Copyright © 2009 Pearson Education, Inc. Order of Operations 1.First, perform all operations within parentheses or other grouping symbols (according to the following order). 2.Next, perform all exponential operations (that is, raising to powers or finding roots). 3.Next, perform all multiplication and divisions from left to right. 4.Finally, perform all additions and subtractions from left to right.

3 Slide 6 - 3 Copyright © 2009 Pearson Education, Inc. Order of Operations Mnemonic Device: Please Excuse (My Dear) (Aunt Sally) Parenthesis, Exponents, (Multiplication & Division), (Addition & Subtraction) **Be careful! Multiplication and division are together (from left to right), and addition and subtraction are together (also from left to right).** 2/(3 – 5 + 1) + 5 2 * 3 = 2/(-2 + 1) + 5 2 * 3in parentheses, 3 – 5 = -2 = 2/(-1) + 5 2 * 3in parentheses, -2 + 1 = -1 = 2/(-1) + 25 * 3 exponent 5 2 = 25 = -2 + 25 * 3dividing 2/(-1) = -2 = -2 + 75multiplying 25 * 3 = 75 = 73addition is all that’s left: -2 + 75 = 73 Order of Operations

4 Slide 6 - 4 Copyright © 2009 Pearson Education, Inc. Definitions Algebra: a generalized form of arithmetic. Variables: letters used to represent numbers Constant: symbol that represents a specific quantity Algebraic expression: a collection of variables, numbers, parentheses, and operation symbols. Examples:

5 Slide 6 - 5 Copyright © 2009 Pearson Education, Inc. Evaluating Expressions When evaluating expressions, we start with an expression or equation with a variable in it. We then substitute the given value every time we see the variable, like so: Evaluate x + 4 for x = 3 Replace x with 3, then follow the order of operations to simplify. x + 4 = 3 + 4 = 7 Evaluate 3x 2 + 4x for x = -2 Replace x with -2, then follow the order of operations to simplify. 3x 2 + 4x = 3(-2) 2 + 4(-2) = 3 * 4 - 8 = 12 - 8 = 4

6 Slide 6 - 6 Copyright © 2009 Pearson Education, Inc. Definitions Terms are parts that are added or subtracted 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. Linear Equations in One Variable

7 Slide 6 - 7 Copyright © 2009 Pearson Education, Inc. Example: Combine Like Terms 8x + 4x = (8 + 4)x = 12x 5x - 6y = 5x - 6y x + 15 - 5x + 9 = (1- 5)x + (15 + 9) = -4x + 24 3x + 2 + 6y - 4 + 7x = (3 + 7)x + 6y + (2 - 4) = 10x + 6y - 2

8 Slide 6 - 8 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 - 9 + 9 = 24 + 9 x = 33 Check: x - 9 = 24 33 - 9 = 24 ? 24 = 24 true

9 Slide 6 - 9 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 19 + 12 = 31 ? 31 = 31 true

10 Slide 6 - 10 Copyright © 2009 Pearson Education, Inc. Solving Equations continued Division Property of Equality If a = b, then Find the solution to the equation 4x = 48.

11 Slide 6 - 11 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. Find the solution to the equation

12 Slide 6 - 12 Copyright © 2009 Pearson Education, Inc. Example: Solving Equations Solve 3x - 4 = 17.

13 Slide 6 - 13 Copyright © 2009 Pearson Education, Inc. Example: Solving Equations Solve 8x + 3 = 6x + 21.

14 Slide 6 - 14 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

15 Slide 6 - 15 Copyright © 2009 Pearson Education, Inc. Example: Solving Equations Solve 21 = 6 + 3(x + 2)

16 Slide 6 - 16 Copyright © 2009 Pearson Education, Inc. 3.3 Formulas Definitions A formula is an equation that typically has a real-life application. To evaluate a formula, substitute the given value for their respective variables and then evaluate using the order of operations.

17 Slide 6 - 17 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 = 300 + 200 P = 500 feet

18 Slide 6 - 18 Copyright © 2009 Pearson Education, Inc. Example The formula for the volume of a cylinder is. Use the formula to find the height of a cylinder with a radius of 6 inches and a volume of 565.49 in 3. The height of the cylinder is 5 inches.

19 Slide 6 - 19 Copyright © 2009 Pearson Education, Inc. Exponential Equations: Carbon Dating Carbon dating is used by scientists to find the age of fossils, bones, and other items. The formula used in carbon dating is If 15 mg of C 14 is present in an animal bone recently excavated, how many milligrams will be present in 4000 years?

20 Slide 6 - 20 Copyright © 2009 Pearson Education, Inc. Exponents View > Scientific 2 [x^y] 7 for 2 7

21 Slide 6 - 21 Copyright © 2009 Pearson Education, Inc. Exponential Equations: Carbon Dating continued In 4000 years, approximately 9.2 mg of the original 15 mg of C 14 will remain. If 15 mg of C 14 is present in an animal bone recently excavated, how many milligrams will be present in 4000 years?

22 Slide 6 - 22 Copyright © 2009 Pearson Education, Inc. Solve for b 2.

23 Slide 6 - 23 Copyright © 2009 Pearson Education, Inc. 3.4 Applications of Linear Equations in One Variable

24 Slide 6 - 24 Copyright © 2009 Pearson Education, Inc. Translating Words to Expressions 2x  3 = 8 Twice a number, decreased by 3 is 8. x – 3 = 4 Three less than a number is 4 x + 7 = 12 Seven more than a number is 12 Mathematical Equation Phrase x  15 = 9x A number decreased by 15 is 9 times the number

25 Slide 6 - 25 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.

26 Slide 6 - 26 Copyright © 2009 Pearson Education, Inc. Example The bill (parts and labor) for the repairs of a car was $496.50. 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 339 + 45h = 496.50

27 Slide 6 - 27 Copyright © 2009 Pearson Education, Inc. Example continued The car was worked on for 3.5 hours.

28 Slide 6 - 28 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?

29 Slide 6 - 29 Copyright © 2009 Pearson Education, Inc. continued, 184 feet of fencing, length 8 feet longer than width Let x = width of region Let x + 8 = length P = 2l + 2w x + 8 x The width of the region is 42 feet and the length is 50 feet.


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