1. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 1.5 Linear Equations, Functions, Zeros, and Applications  Solve linear equations.  Solve applied problems using linear models.  Find zeros of linear functions.

2. Slide 1.5 - 2 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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

3. Slide 1.5 - 3 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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.

4. Slide 1.5 - 4 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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.

5. Slide 1.5 - 5 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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.

6. Slide 1.5 - 6 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Solve: Solution: Start by multiplying both sides of the equation by the LCD to clear the equation of fractions.

7. Slide 1.5 - 7 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) Check: The solution is TRUE

8. Slide 1.5 - 8 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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.

9. Slide 1.5 - 9 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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 (–∞, ∞).

10. Slide 1.5 - 10 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Applications Using Linear Models Mathematical techniques are used to answer questions arising from real-world situations. Linear equations and functions model many of these situations.

11. Slide 1.5 - 11 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Five Steps for Problem Solving 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.

12. Slide 1.5 - 12 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley The Motion Formula The distance d traveled by an object moving at rate r in time t is given by d = rt.

13. Slide 1.5 - 13 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example 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?

14. Slide 1.5 - 14 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 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.

15. Slide 1.5 - 15 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 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

16. Slide 1.5 - 16 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 500t = 302(t + 1) 500t = 302t + 302 198t = 302 t = 302/198 ≈ 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.

17. Slide 1.5 - 17 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Simple-Interest Formula I = Prt I = the simple interest ($) P = the principal ($) r = the interest rate (%) t = time (years)

18. Slide 1.5 - 18 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example 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?

19. Slide 1.5 - 19 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 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

20. Slide 1.5 - 20 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 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.

21. Slide 1.5 - 21 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 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.

22. Slide 1.5 - 22 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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.

23. Slide 1.5 - 23 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Find the zero of f (x) = 5x  9. Algebraic Solution: 5x  9 = 0 5x = 9 x = 1.8 Visualizing the Solution: The intercept of the graph is (9/5, 0) or (1.8, 0).