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# Chapter 3: Systems of Linear Equations

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1. Chapter 3: Systems of Linear Equations

2. 3.1 Solving Systems of Linear Equations by Graphing

3. Two or more linear equations form a system of linear equations.Example: • A solution to a system of linear equationsis an ordered pair that is a solution to each individual linear equation.(–2, –4) is a solution to the system above because it is a solution to each equation.

4. A solution to a system of two linear equations may be interpreted graphically as a point of intersection between the two lines.

5. Two lines drawn in a rectangular coordinate system may intersect at exactly one point, no point, or infinitely many points (the lines are coinciding).

6. A consistent system of linear equations has one or more solutions. • An inconsistent system has no solution. • A dependent system has infinitely many solutions (both equations in the system represent the same line). • An independent system is one in which the two equations represent different lines.

7. To solve a system of linear equations by graphing, graph both equations and finding the point(s) of intersection. It can be helpful to first write the equations in slope-intercept form. • Different slopes: lines are different and nonparallel, thus must cross in exactly one point • Same slopes, different y-intercepts: lines are parallel and do not intersect • Same slopes, same y-intercepts: equations represent the same line

8. Exercise 1 Which point is a solution to the system? 10x + 9y = 154 –x + 10y = 50 A) (10, 10) B) (10, 6) C) (6, 10) D) (6, 6)

9. Exercise 2 Which of the following describes the system of equations whose graph is shown? A) Consistent and Independent B) Consistent and Dependent C) Inconsistent and Independent D) Inconsistent and Dependent

10. Exercise 3 Solve the system using the graphing method. x + y = 13 3x + y = 35 A) (12, –1) B) (11, 2) C) (2, 11) D) No solution

11. The substitution method is another technique used to solve a system of equations. • This method can be used when the system includes nonlinear equations.

12. Solving a System of Equations by the Substitution Method • Isolate one of the variables from one equation. • Substitute the quantity found in step 1 into the other equation. • Solve the resulting equation. • Substitute the value found in in step 3 back into the equation in step 1 to find the value of the remaining variable. • Check the solution in both equations, and write the answer as an ordered pair.

13. ExampleSolve the system by using the substitution method. 3x – 2y = –7 6x + y = 6 Step 1: Solve the second equation for y.

14. Example (continued)Solve the system by using the substitution method. 3x – 2y = –76x + y = 6 Step 2: Substitute the quantity –6x + 6 for y in the other equation.

15. Example (continued)Solve the system by using the substitution method. 3x – 2y = –76x + y = 6 Step 3: Solve for x.

16. Example (continued)Solve the system by using the substitution method. 3x – 2y = –76x + y = 6 Step 4: Substitute x = into the expression y = –6x + 6.

17. Example (continued)Solve the system by using the substitution method. 3x – 2y = –76x + y = 6 Step 5: Check the ordered pair in each original equation. The solution is .

18. When using the substitution method to solve a system of linear equations, if the equation in step 3 reduces to a contradiction, then the system has no solution and is inconsistent. • If the equation in step 3 reduces to an identity, the system is dependent and the solution may be represented by the set of all solutions to either equation in the system.

19. Exercise 4 Solve the system using the substitution method. y = –63 – 7x 3x + 7y = –73 A) (8, –8) B) (–8, –7) C) (–7, –8) D) (–7, 9)

20. Exercise 5 Solve the system using the substitution method. y = 7x + 8 –3y + 21x = –6   A) (8, –6) B) (0, 8) C) There are infinitely many solutions.  D) There is no solution.

21. Exercise 6 Solve the system using the substitution method. 6x + 10y = 6 30y = 18 – 18x A) (–10, 7) B) (7, –10) C) {(x, y) | 6x + 10y = 6} D) There is no solution.

22. Solving a System of Equations by the Addition Method • Write both equations in standard form Ax + By = C. • Clear fractions or decimals (optional). • Multiply one or both equations by nonzero constants to create opposite coefficients for one of the variables. • Add the equations from step 3 to eliminate one variable. • Solve for the remaining variable. • Substitute the known value found in step 5 into one of the original equations to solve for the other variable. • Check the ordered pair in both equations.

23. ExampleSolve the system by using the addition method. 4x + 5y = 23x = 1 – 4y Step 1: Write both equations in standard form. 4x + 5y = 2 3x + 4y = 1 Step 2: There are no fractions or decimals.

24. Example (continued)Solve the system by using the addition method. 4x + 5y = 23x = 1 – 4y Step 3: Multiply the first equation by 3. Multiply the second equation by –4. 4x + 5y = 2 12x + 15y = 6 3x + 4y = 1 –12x – 16y = –4 Step 4: Add the equations. 12x + 15y = 6 –12x – 16y = –4 –y = 2

25. Example (continued)Solve the system by using the addition method. 4x + 5y = 23x = 1 – 4y Step 5: Solve for y. –y = 2 y = –2 Step 6: Substitute y = –2 back into one of the original equations and solve for x. 4x + 5y = 2 4x + 5(–2) = 2 4x – 10 = 2 4x = 12 x = 3

26. Example (continued)Solve the system by using the addition method. 4x + 5y = 23x = 1 – 4y Step 7: Check the ordered pair (3,–2) in both original equations. The solution is (3,–2).

27. If both variables are eliminated when the equations are added, the two original equations are equivalent and the system is dependent. The solution may be represented by the set of all solutions to either equation in the system. • If the equations reduce to a contradiction when added, the system has no solution. The system is inconsistent and the two equations represent parallel lines.

28. Exercise 7 Which step would successfully eliminate y from the system below? 6x + 8y = 27 x + 24y = 54   A) Add the 2 equations.  B) Multiply the first equation by 3 and add the equations.  C) Multiply the second equation by 6 and add the equations.  D) Multiply the first equation by –3 and add the equations.

29. Exercise 8 Solve the system using the addition method. 5x + y = 33 7x – y = 27 A) (5, 8) B) (2, 9) C) (8, 5) D) There is no solution.

30. Exercise 9 Solve the system of equations using the addition method. 10x – 8y = 0 20x = 16y + 0 A) {(x, y): 10x – 8y = 0} B) (0, 0) C) (10, 8) D) There is no solution.

31. When an application has more than one unknown, it may be necessary to use multiple variables. • When two variables are present, the goal is to set up a system of two independent equations. • To solve the system, use the graphing method, the substitution method, or the addition method.

32. There are several types of application problems that can be solved using a system of linear equations in two variables: • Cost applications • Mixture applications • Applications involving principal and interest • Applications involving distance, rate, and time • Geometry applications

33. Steps to Solve Applications • Label two variables. • Construct two equations in words. • Write two equations. • Solve the system. • Write the answer.

34. Exercise 10 At one store, 5 pairs of jeans and 2 sweatshirts costs \$166, while 3 pairs of jeans and 4 sweatshirts costs \$164. Find the cost of one sweatshirt. A) \$23 B) \$24 C) \$20 D) \$26

35. Exercise 11 A plane can fly 1,488 miles with the wind in 3 hours, while it takes the same plane 4 hours to fly the same distance against the wind. Find the speed of the plane in still air and the speed of the wind.   A) Plane: 496 mph, Wind: 62 mph  B) Plane: 992 mph, Wind: 124 mph C) Plane: 372 mph, Wind: 124 mph D) Plane: 434 mph, Wind: 62 mph

36. Exercise 12 A new sports car, the Revolution, can be leased for \$3,000 down and \$300 per month, so that the cost of leasing is given by y = 3000 + 300x, where x is the length of the lease in months. The same dealership offers the Yachtsman, a luxury car, for \$5,000 down and \$260 per month. After how many months would the total cost of leasing the two cars be equal? A) 12 months B) 36 months C) 45 months D) 50 months

37. A linear equation in three variables can be written in the form Ax + By + Cz = D, where A, B, and C are not all zero. • Solutions to such an equation are ordered triples of the form (x, y, z) that satisfy the equation.

38. The set of all ordered triples that are solutions to a linear equation in three variables may be represented graphically by a plane in space.

39. A solution to a system of linear equations in three variables is an ordered triple that satisfies each equation. • Geometrically, a solution is a point of intersection of the planes represented by the equations in the system. • A system of linear equations in three variables may have one unique solution, infinitely many solutions, or no solution.

40. One unique solution (planes intersect at one point) • The system is consistent. • The system is independent.

41. No solution(the three planes do not all intersect) • The system is inconsistent. • The system is independent.

42. Infinitely many solutions(planes intersect at infinitely many points) • The system is consistent. • The system is dependent.

43. Solving a System of Three Linear Equations in Three Variables • Write each equation in standard form Ax + By + Cz = D. It is often helpful to label the equations. • Choose a pair of equations and eliminate one of the variables by using the addition method. • Choose a different pair of equations and eliminate the same variable.

44. Solving a System of Three Linear Equations in Three Variables • After step 3 there should be two equations in two variables. Solve this system using the addition method or the substitution method. • Substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable. Solve for the third variable. • Check the ordered triple in each of the original equations.

45. ExampleSolve the system. Step 1:The equations are already in standard form. Label the equations. A B C