1. Section 1.2 Solving Equations Using a Graphing Utility; Linear and Rational Equations Copyright © 2013 Pearson Education, Inc. All rights reserved

2. An equation in one variable is a statement in which two expressions, at least one containing the variable, are equal. The expressions are called the sides of the equation. The admissible values of the variable, if any, that result in a true statement are called solution, or roots, of the equation. To solve an equation means to find all the solutions of the equation. Equations in One Variable Copyright © 2013 Pearson Education, Inc. All rights reserved

3. Examples of an equation in one variable: Equations in One Variable x + 5 = 9 is true when x = 4. 4 is a solution of the equation or we say that 4 satisfies the equation. We write the solution in set notation, this is called the solution set of the equation. An equation that is satisfied for every choice of the variable for which both sides are defined is called an identity. Copyright © 2013 Pearson Education, Inc. All rights reserved

4. Solve Equations Using a Graphing Utility Copyright © 2013 Pearson Education, Inc. All rights reserved

5. Using ZERO (or ROOT) to Approximate Solutions of an Equation Find the solution(s) of the equation x 3 – x + 1 = 0. Round answers to two decimal places. The solutions of the equation x 3 – x + 1 = 0 are the same as the x-intercepts of the graph of Y 1 = x 3 – x + 1. There appears to be one x-intercept (solution) between –2 and –1.

6. Copyright © 2013 Pearson Education, Inc. All rights reserved Using ZERO (or ROOT) to Approximate Solutions of an Equation Using the ZERO (or ROOT) feature of a graphing utility, we determine that the x- intercept, and thus the solution to the equation, is x = –1.32 rounded to two decimal places.

7. Copyright © 2013 Pearson Education, Inc. All rights reserved Using INTERSECT to Approximate Solutions of an Equation Find the solution(s) to the equation 4x 4 – 3 = 2x + 1. Round answers to two decimal places. Begin by graphing each side of the equation: Y 1 = 4x 4 – 3 and Y 2 = 2x + 1.

8. Copyright © 2013 Pearson Education, Inc. All rights reserved Using INTERSECT to Approximate Solutions of an Equation At a point of intersection of the graphs, the value of the y-coordinate is the same for Y 1 and Y 2. Thus, the x-coordinate of the point of intersection represents a solution to the equation.

9. Copyright © 2013 Pearson Education, Inc. All rights reserved Using INTERSECT to Approximate Solutions of an Equation The INTERSECT feature on a graphing utility determines a point of intersection of the graphs. Using this feature, we find that the graphs intersect at (–0.87, –0.73) and (1.12, 3.23) rounded to two decimal places.

10. Try this: P. 106 # 20 Copyright © 2013 Pearson Education, Inc. All rights reserved

12. One method for solving equations algebraically requires that a series of equivalent equations be developed from the original equation until an obvious solution results. Solving Equations Algebraically

13. Copyright © 2013 Pearson Education, Inc. All rights reserved

19. Try this: p. 106 # 39 Copyright © 2013 Pearson Education, Inc. All rights reserved

20. Solve Rational Equations Copyright © 2013 Pearson Education, Inc. All rights reserved

21. A rational equation is an equation that contains a rational expression. To solve a rational equation, multiply both sides of the equation by the least common multiple of the denominators of the rational expressions that make up the rational equation. Copyright © 2013 Pearson Education, Inc. All rights reserved

22. Solving a Rational Equation

23. no solution Copyright © 2013 Pearson Education, Inc. All rights reserved

24. Try This: P. 107 # 63 Copyright © 2013 Pearson Education, Inc. All rights reserved

25. Sometimes the process of creating equivalent equations leads to apparent solutions that are not solutions of the original equation. These are called extraneous solutions. Extraneous Solutions Copyright © 2013 Pearson Education, Inc. All rights reserved

28. A total of $16,000 is invested, some in stocks and some in bonds. If the amount invested in bonds is one fourth that invested in stocks, how much is invested in each category? We are being asked to find the amount of two investments. These amounts total $16,000. Copyright © 2013 Pearson Education, Inc. All rights reserved

29. A total of $16,000 is invested, some in stocks and some in bonds. If the amount invested in bonds is one fourth that invested in stocks, how much is invested in each category? If x equals the amount invested in stocks, then the rest of the money, 16,000 – x, is the amount invested in bonds. Copyright © 2013 Pearson Education, Inc. All rights reserved

30. A total of $16,000 is invested, some in stocks and some in bonds. If the amount invested in bonds is one fourth that invested in stocks, how much is invested in each category? We also know that: Total amount invest in bonds is one fourth that in stocks Copyright © 2013 Pearson Education, Inc. All rights reserved

31. A total of $16,000 is invested, some in stocks and some in bonds. If the amount invested in bonds is one fourth that invested in stocks, how much is invested in each category? So $3200 is invested in stocks and $16000 – $3200 = $12,800 is invested in bonds. Copyright © 2013 Pearson Education, Inc. All rights reserved

32. A total of $16,000 is invested, some in stocks and some in bonds. If the amount invested in bonds is one fourth that invested in stocks, how much is invested in each category? The total invested is $3200 + $12,800 = $16000 and the amount of bonds, $3200 is one fourth of that of stocks, $12,800. Copyright © 2013 Pearson Education, Inc. All rights reserved

33. Try This P. 107 # 96 Copyright © 2013 Pearson Education, Inc. All rights reserved

34. Homework P. 106 # 15 – 75 (x5), 95, 99 Copyright © 2013 Pearson Education, Inc. All rights reserved

35. Andy grossed $440 one week by working 50 hours. His employer pays time-and-a-half for all hours worked in excess of 40 hours. What is Andy’s hourly wage? We are looking for an hourly wage. Our answer will be expressed in dollars per hour. Copyright © 2013 Pearson Education, Inc. All rights reserved

36. Andy grossed $440 one week by working 50 hours. His employer pays time-and-a-half for all hours worked in excess of 40 hours. What is Andy’s hourly wage? Let x represent the regularly wage; x is measured in dollars per hour. Copyright © 2013 Pearson Education, Inc. All rights reserved

37. Andy grossed $440 one week by working 50 hours. His employer pays time-and-a-half for all hours worked in excess of 40 hours. What is Andy’s hourly wage? The sum of regular salary plus overtime salary wil equal $440. From the table, 40x + 15x = 440. Copyright © 2013 Pearson Education, Inc. All rights reserved

38. Andy grossed $440 one week by working 50 hours. His employer pays time-and-a-half for all hours worked in excess of 40 hours. What is Andy’s hourly wage? 40x + 15x = 440 55x = 440 x = 8 Andy’s regularly wage is $8.00 per hour. Copyright © 2013 Pearson Education, Inc. All rights reserved

39. Andy grossed $440 one week by working 50 hours. His employer pays time-and-a-half for all hours worked in excess of 40 hours. What is Andy’s hourly wage? Forty hours yields a salary of 40(8.00) = $320.00 and 10 hours overtime yields a salary of 10(1.5)(8.00) = $120.00 for a total of $440.00. Copyright © 2013 Pearson Education, Inc. All rights reserved