Presentation is loading. Please wait.

Presentation is loading. Please wait.

 SOLVE LINEAR EQUATIONS.  SOLVE APPLIED PROBLEMS USING LINEAR MODELS.  FIND ZEROS OF LINEAR FUNCTIONS. Copyright © 2012 Pearson Education, Inc. Publishing.

Similar presentations


Presentation on theme: " SOLVE LINEAR EQUATIONS.  SOLVE APPLIED PROBLEMS USING LINEAR MODELS.  FIND ZEROS OF LINEAR FUNCTIONS. Copyright © 2012 Pearson Education, Inc. Publishing."— Presentation transcript:

1  SOLVE LINEAR EQUATIONS.  SOLVE APPLIED PROBLEMS USING LINEAR MODELS.  FIND ZEROS OF LINEAR FUNCTIONS. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 1.5 Linear Equations, Functions, Zeros, and Applications

2 Equations and Solutions Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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 Linear Equations Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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 Equation-Solving Principles Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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. Equations that have the same solution set are equivalent equations.

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

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

7 Example - Special Case Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Solve: Some equations have no solution.

8 Example - Special Case Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Solve: 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 (–∞, ∞).

9 Five Steps for Problem Solving Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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. Applications Using Linear Models

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

11 Example Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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?

12 Example (continued) Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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.

13 Example (continued) Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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

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

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

16 Example Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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?

17 Example (continued) Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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

18 Example (continued) Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 2. Translate. The total amoun t 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.

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

20 Zeros of Linear Functions Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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.

21 Example Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Find the zero of f (x) = 5x  9.


Download ppt " SOLVE LINEAR EQUATIONS.  SOLVE APPLIED PROBLEMS USING LINEAR MODELS.  FIND ZEROS OF LINEAR FUNCTIONS. Copyright © 2012 Pearson Education, Inc. Publishing."

Similar presentations


Ads by Google