2. Copyright © 2009 Pearson Education, Inc. CHAPTER 1: Graphs, Functions, and Models 1.1 Introduction to Graphing 1.2 Functions and Graphs 1.3 Linear Functions, Slope, and Applications 1.4 Equations of Lines and Modeling 1.5 Linear Equations, Functions, Zeros and Applications 1.6 Solving Linear Inequalities

3. Copyright © 2009 Pearson Education, Inc. 1.5 Linear Equations, Functions, Zeros, and Applications  Solve linear equations.  Solve applied problems using linear models.  Find zeros of linear functions.  Solve a formula for a given variable.

4. Slide 1.5 - 4 Copyright © 2009 Pearson Education, Inc. Equations and Solutions An equation is a statement that two expressions are equal. To solve an equation in one variable is to find all the values of the variable that make the equation true. Each of these numbers is a solution of the equation. The set of all solutions of an equation is its solution set. Some examples of equations in one variable are

5. Slide 1.5 - 5 Copyright © 2009 Pearson Education, Inc. Linear Equations A linear equation in one variable is an equation that can be expressed in the form mx + b = 0, where m and b are real numbers and m ≠ 0.

6. Slide 1.5 - 6 Copyright © 2009 Pearson Education, Inc. Equivalent Equations Equations that have the same solution set are equivalent equations. For example, 2x + 3 = 5 and x = 1 are equivalent equations because 1 is the solution of each equation. On the other hand, x 2 – 3x + 2 = 0 and x = 1 are not equivalent equations because 1 and 2 are both solutions of x 2 – 3x + 2 = 0 but 2 is not a solution of x = 1.

7. Slide 1.5 - 7 Copyright © 2009 Pearson Education, Inc. Equation-Solving Principles For any real numbers a, b, and c: The Addition Principle: If a = b is true, then a + c = b + c is true. The Multiplication Principle: If a = b is true, then ac = bc is true.

8. Slide 1.5 - 8 Copyright © 2009 Pearson Education, Inc. Example Solve: Solution: Start by multiplying both sides of the equation by the LCD to clear the equation of fractions.

9. Slide 1.5 - 9 Copyright © 2009 Pearson Education, Inc. Example (continued) Check: The solution is TRUE

10. Slide 1.5 - 10 Copyright © 2009 Pearson Education, Inc. Example - Special Case Solve: Solution: Some equations have no solution. No matter what number we substitute for x, we get a false sentence. Thus, the equation has no solution.

11. Slide 1.5 - 11 Copyright © 2009 Pearson Education, Inc. Example - Special Case Solve: Solution: There are some equations for which any real number is a solution. Replacing x with any real number gives a true sentence. Thus any real number is a solution. The equation has infinitely many solutions. The solution set is the set of real numbers, {x | x is a real number}, or (–∞, ∞).

12. Slide 1.5 - 12 Copyright © 2009 Pearson Education, Inc. Applications Using Linear Models (optional) Mathematical techniques are used to answer questions arising from real-world situations. Linear equations and functions model many of these situations.

13. Slide 1.5 - 13 Copyright © 2009 Pearson Education, Inc. Five Steps for Problem Solving (optional) 1.Familiarize yourself with the problem situation. Make a drawing Find further information Assign variablesOrganize into a chart or table Write a list Guess or estimate the answer 2.Translate to mathematical language or symbolism. 3.Carry out some type of mathematical manipulation. 4.Check to see whether your possible solution actually fits the problem situation. 5.State the answer clearly using a complete sentence.

14. Slide 1.5 - 14 Copyright © 2009 Pearson Education, Inc. The Motion Formula (optional) The distance d traveled by an object moving at rate r in time t is given by d = rt.

15. Slide 1.5 - 15 Copyright © 2009 Pearson Education, Inc. Example (optional) America West Airlines’ fleet includes Boeing 737-200’s,each with a cruising speed of 500 mph, and Bombardier deHavilland Dash 8-200’s, each with a cruising speed of 302 mph (Source: America West Airlines). Suppose that a Dash 8- 200 takes off and travels at its cruising speed. One hour later, a 737-200 takes off and follows the same route, traveling at its cruising speed. How long will it take the 737-200 to overtake the Dash 8-200?

16. Slide 1.5 - 16 Copyright © 2009 Pearson Education, Inc. Example (continued) (optional) 1.Familiarize. Make a drawing showing both the known and unknown information. Let t = the time, in hours, that the 737-200 travels before it overtakes the Dash 8-200. Therefore, the Dash 8-200 will travel t + 1 hours before being overtaken. The planes will travel the same distance, d.

17. Slide 1.5 - 17 Copyright © 2009 Pearson Education, Inc. Example (continued) (optional) We can organize the information in a table as follows. 2.Translate. Using the formula d = rt, we get two expressions for d: d = 500t and d = 302(t + 1). Since the distance are the same, the equation is: 500t = 302(t + 1) DistanceRateTime 737-200d500t Dash 8-200d302t + 1 d = r t

18. Slide 1.5 - 18 Copyright © 2009 Pearson Education, Inc. Example (continued) (optional) 500t = 302(t + 1) 500t = 302t + 302 198t = 302 t ≈ 1.53 4. Check. If the 737-200 travels about 1.53 hours, it travels about 500(1.53) ≈ 765 mi; and the Dash 8-200 travels about 1.53 + 1, or 2.53 hours and travels about 302(2.53) ≈ 764.06 mi, the answer checks. (Remember that we rounded the value of t). 5. State. About 1.53 hours after the 737-200 has taken off, it will overtake the Dash 8-200. 3. Carry out. We solve the equation.

19. Slide 1.5 - 19 Copyright © 2009 Pearson Education, Inc. Simple-Interest Formula (optional) I = Prt I = the simple interest ($) P = the principal ($) r = the interest rate (%) t = time (years)

20. Slide 1.5 - 20 Copyright © 2009 Pearson Education, Inc. Example (optional) Jared’s two student loans total $12,000. One loan is at 5% simple interest and the other is at 8%. After 1 year, Jared owes $750 in interest. What is the amount of each loan?

21. Slide 1.5 - 21 Copyright © 2009 Pearson Education, Inc. Example (continued) (optional) Solution: 1. Familiarize. We let x = the amount borrowed at 5% interest. Then the remainder is $12,000 – x, borrowed at 8% interest. Amount Borrowed Interest Rate TimeAmount of Interest 5% loanx0.0510.05x 8% loan12,000 – x0.0810.08(12,000 – x) Total12,000750

22. Slide 1.5 - 22 Copyright © 2009 Pearson Education, Inc. Example (continued) (optional) 2. Translate. The total amount of interest on the two loans is $750. Thus we write the following equation. 0.05x + 0.08(12,000  x) = 750 3. Carry out. We solve the equation. 0.05x + 0.08(12,000  x) = 750 0.05x + 960  0.08x = 750  0.03x + 960 = 750  0.03x =  210 x = 7000 If x = 7000, then 12,000  7000 = 5000.

23. Slide 1.5 - 23 Copyright © 2009 Pearson Education, Inc. Example (continued) (optional) 4. Check. The interest on $7000 at 5% for 1 yr is $7000(0.05)(1), or $350. The interest on $5000 at 8% for 1 yr is $5000(0.08)(1) or $400. Since $350 + $400 = $750, the answer checks. 5. State. Jared borrowed $7000 at 5% interest and $5000 at 8% interest.

24. Slide 1.5 - 24 Copyright © 2009 Pearson Education, Inc. Zeros of Linear Functions An input c of a function f is called a zero of the function, if the output for the function is 0 when the input is c. That is, c is a zero of f if f (c) = 0. A linear function f (x) = mx + b, with m  0, has exactly one zero.

25. Slide 1.5 - 25 Copyright © 2009 Pearson Education, Inc. Example Find the zero of f (x) = 5x  9. Algebraic Solution: 5x  9 = 0 5x = 9 x = 1.8 Using a table in ASK mode we can check the solution. Enter y = 5x – 9 into the equation editor, then enter x = 1.8 into the table and it yields y = 0. That means the zero is 1.8.

26. Slide 1.5 - 26 Copyright © 2009 Pearson Education, Inc. Example (continued) Find the zero of f (x) = 5x  9. Graphical Solution or Zero Method: Graph y = 5x – 9. Use the ZERO feature from the CALC menu. The x-intercept is 1.8, so the zero of f (x) = 5x – 9 is 1.8.

27. Slide 1.5 - 27 Copyright © 2009 Pearson Education, Inc. Formulas A formula is an equation that can be used to model a situation. We have used the motion formula, d = r t. We have used the simple-interest formula, I = Prt. We can use the equation-solving principles to solve a formula for a given variable.

28. Slide 1.5 - 28 Copyright © 2009 Pearson Education, Inc. Example Solve P = 2l + 2w for l. Solution: The formulacan be used to determine a rectangle’s length if we know its perimeter and width.