College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

1. College Algebra Fifth Edition James StewartLothar RedlinSaleem Watson

2. Systems of Equations and Inequalities 6

3. Chapter Overview • We have already seen how a real-world situation can be modeled by an equation. • However, Many real-world situations have too many variables to be modeled by a singleequation. • For example, weather depends on many variables—temperature, wind speed, air pressure, humidity, and so on.

4. Chapter Overview • So, to model (and forecast) the weather, scientists use many equations—each having many variables. • Such systems of equations work togetherto describe the weather.

5. Chapter Overview • Systems of equations with hundreds, or even thousands, of variables are also used extensively in the air travel and telecommunications industries to: • Establish consistent airline schedules. • Find efficient routing for telephone calls.

6. Systems of Equations 6.1

7. Systems of Equations • Here, we study how to solve systems of two equations in two unknowns. • We learn three different methods of solving such systems: • By substitution • By elimination • Graphically

8. Systems of Equations and their Solutions

9. Systems of Equations and their Solutions • A system of equationsis a set of equations that involve the same variables. • A solution of a system is an assignment of values for the variables that makes eachequation in the system true. • To solvea system means to find all solutions of the system.

10. Systems of Equations and their Solutions • Here is an example of a system of two equations in two variables:

11. Systems of Equations and their Solutions • We can check that x = 3 and y = 1 is a solution of this system. • The solution can also be written as the ordered pair (3, 1).

12. Systems of Equations and their Solutions • Note that the graphs of Equations 1 and 2 are lines. • As the solution (3, 1) satisfies each equation, the point (3, 1) lies on each line. • So, it is the point of intersection of the two lines.

13. Substitution Method

14. Substitution Method • In the substitution method,we start with one equation in the system and solve for one variable in terms of the other variable. • Solve for one variable. • Substitute. • Back-substitute.

15. Step 1 • Solve for one variable. • Choose one equation and solve for one variable in terms of the other variable.

16. Step 2 • Substitute. • Substitute the expression you found in step 1 into the other equation to get an equation in one variable. • Then, solve for that variable.

17. Step 3 • Back-substitute. • Substitute the value you found in step 2 back into the expression found in step 1 to solve for the remaining variable.

18. E.g. 1—Substitution Method • Find all solutions of the system. • We solve for y in the first equation.y = 1 – 2x

19. E.g. 1—Substitution Method • Now, we substitute for y in the second equation and solve for x: 3x + 4(1 – 2x) = 14 3x + 4 – 8x = 14 –5x + 4 = 14 –5x = 10x = –2

20. E.g. 1—Substitution Method • Next, we back-substitute x = –2 into the equation y = 1 – 2x:y = 1 – 2(–2) = 5 • Thus, x = –2 and y = 5. • So, the solution is the ordered pair (–2, 5).

21. E.g. 1—Substitution Method • The figure shows that the graphs of the two equations intersect at the point (–2, 5).

22. E.g. 2—Substitution Method • Find all solutions of the system. • We start by solving for y in the second equation.y = 3x – 10

23. E.g. 2—Substitution Method • Next, we substitute for y in the first equation and solve for x:x2 + (3x – 10)2 = 100x2 + (9x2 – 60x + 100) = 100 10x2 – 60x = 0 10x(x – 6) = 0x = 0 or x = 6

24. E.g. 2—Substitution Method • Now, we back-substitute these values of x into the equation y = 3x – 10. • For x = 0: y = 3(0) – 10 = –10 • For x = 6: y = 3(6) – 10 = 8 • So, we have two solutions: (0, –10) and (6, 8).

25. E.g. 2—Substitution Method • The graph of the first equation is a circle. • That of the second equation is a line. • The graphs intersect at the two points (0, –10) and (6, 8).

26. Elimination Method

27. Elimination Method • To solve a system using the elimination method, we try to combine the equations using sums or differences so as to eliminate one of the variables. • Adjust the coefficients. • Add the equations. • Back-substitute.

28. Step 1 • Adjust the coefficients. • Multiply one or more of the equations by appropriate numbers so that the coefficient of one variable in one equation is the negative of its coefficient in the other equation.

29. Step 2 • Add the equations. • Add the two equations to eliminate one variable. • Then, solve for the remaining variable.

30. Step 3 • Back-substitute. • Substitute the value you found in step 2 back into one of the original equations. • Then, solve for the remaining variable.

31. E.g. 3—Elimination Method • Find all solutions of the system. • The coefficients of the y-terms are negatives of each other. • So, we can add the equations to eliminate y.

32. E.g. 3—Elimination Method • Now, we back-substitute x = 4 into one of the original equations and solve for y. • Let’s choose the second equation because it looks simpler.

33. E.g. 3—Elimination Method • x – 2y = 24 – 2y = 2 –2y = –2y = 1 • The solution is (4, 1).

34. E.g. 3—Elimination Method • The figure shows that the graphs of the equations in the system intersect at the point (4, 1).

35. E.g. 4—Elimination Method • Find all solutions of the system. • We choose to eliminate the x-term. • So, we multiply the first equation by 5 and the second equation by –3. • Then, we add the two equations and solve for y.

36. E.g. 4—Elimination Method • Now, we back-substitute y = –11 into one of the original equations (say, 3x2 + 2y = 26) and solve for x.

37. E.g. 4—Elimination Method • 3x2 + 2(–11) = 26 3x2 = 48x2 = 16x = –4 or x = 4 • So, we have two solutions: (–4, –11) and (4, –11)

38. E.g. 4—Elimination Method • The graphs of both equations are parabolas. • The graphs intersect at the two points (–4, –11) and (4, –11).

39. Graphical Method

40. Graphical Method • In the graphical method,we use a graphing device to solve the system of equations. • Note that, with many graphing devices, any equation must first be expressed in terms of one or more functions of the form y = f(x) before we can use the calculator to graph it.

41. Graphical Method • Not all equations can be readily expressed in this way. • So, not all systems can be solved by this method.

42. Graphical Method • We proceed as follows. • Graph each equation. • Find the intersection points.

43. Step 1 • Graph each equation. • Express each equation in a form suitable for the graphing calculator by solving for y as a function of x. • Graph the equations on the same screen.

44. Step 2 • Find the intersection points. • The solutions are the x- and y-coordinates of the points of intersection.

45. Graphical Method • It may be more convenient to solve for x in terms of y in the equations. • In that case, in step 1, graph x as a function of y instead.

46. E.g. 5—Graphical Method • Find all solutions of the system. • Solving for y in terms of x, we get the equivalent system

47. E.g. 5—Graphical Method • The figure shows that the graphs of these equations intersect at two points. • Zooming in, we see that the solutions are(–1, –1) and (3, 7).

48. E.g. 6—Solving a System of Equations Graphically • Find all solutions of the system, correct to one decimal place. • The graph of the first equation is a circle and the second a parabola. • To graph the circle on a graphing calculator, we must first solve for y in terms of x (Section 2.3).

49. E.g. 6—Solving a System of Equations Graphically • To graph the circle, we must graph both functions:

50. E.g. 6—Solving a System of Equations Graphically • In the figure, the graph of the circle is shown in red and the parabola in blue.