1. Topic 6 : Systems 6.1 System of linear equations

2. 5-Minute Check 1 A.D = {–4, –2, 0, 1, 2}, R = {–4, 0,1}; yes B.D = {0, 1, 2}, R = {0, 1}; yes C.D = {–4, 0, 1}, R = {–4, –2, 0, 1, 2}; no D.D = {–2, –4}; R = {–4, 0, 1}; yes Find the domain and range of the relation {(–4, 1), (0, 0), (1, –4), (2, 0), (–2, 0)}. Determine whether the relation is a function. Opening Routine

3. 5-Minute Check 2 A.28 B.12 C.–12 D.–16 Find the value of f(4) for f(x) = 8 – x – x 2.

4. 5-Minute Check 5 A.15 B.16 C.17 D.18 The Math Club is using the prediction equation y = 1.25x + 10 to estimate the number of members it will have, where x represents the number of years the club has been in existence. About how many members does the club expect to have in its fifth year?

5. 5-Minute Check 6 A.absolute value B.linear C.piecewise-defined D.quadratic Identify the type of function represented by the equation y = 4x 2 + 6.

6. Vocabulary break-even point system of equations consistent inconsistent independent dependent substitution method elimination method

7. Example 1 Solve by Using a Table Solve the system of equations by completing a table. x + y = 3 –2x + y = –6 x + y =3 y =–x + 3 Solve for y in each equation. –2x + y =–6 y =2x – 6

8. Example 1 Solve by Using a Table Answer: The solution to the system is (3, 0). Use a table to find the solution that satisfies both equations.

9. Example 1 What is the solution of the system of equations? x + y = 2 x – 3y = –6 A.(1, 1) B.(0, 2) C.(2, 0) D.(–4, 6)

10. Example 2 Solve the system of equations by graphing. x – 2y = 0 x + y = 6 Solve by Graphing The graphs appear to intersect at (4, 2). Write each equation in slope-intercept form.

11. Example 2 Check Substitute the coordinates into each equation. Solve by Graphing x – 2y= 0x + y=6Original equations 4 – 2(2)= 04 + 2=6Replace x with 4 and y with 2. ?? 0=06=6Simplify. Answer: The solution of the system is (4, 2).

12. Example 2 Which graph shows the solution to the system of equations below? x + 3y=7 x – y=3 A.C. B.D.

13. Example 3 A. Graph the system of equations and describe it as consistent and independent, consistent and dependent, or inconsistent. x – y = 5 x + 2y = –4 Classify Systems Write each equation in slope-intercept form.

14. Example 3 Answer: Classify Systems The graphs of the equations intersect at (2, –3). Since there is one solution to this system, this system is consistent and independent.

15. Example 3 B. Graph the system of equations and describe it as consistent and independent, consistent and dependent, or inconsistent. 9x – 6y = –6 6x – 4y = –4 Classify Systems Write each equation in slope-intercept form. Since the equations are equivalent, their graphs are the same line.

16. Example 3 Answer: Classify Systems Any ordered pair representing a point on that line will satisfy both equations. So, there are infinitely many solutions. This system is consistent and dependent.

17. Example 3 C. Graph the system of equations and describe it as consistent and independent, consistent and dependent, or inconsistent. 15x – 6y = 0 5x – 2y = 10 Classify Systems Write each equation in slope-intercept form.

18. Example 3 Answer : Classify Systems The lines do not intersect. Their graphs are parallel lines. So, there are no solutions that satisfy both equations. This system is inconsistent.

19. Example 3 D. Graph the system of equations and describe it as consistent and independent, consistent and dependent, or inconsistent. f(x) = –0.5x + 2 g(x) = –0.5x + 2 h(x) = 0.5x + 2 Classify Systems

20. Example 3 Answer: Classify Systems f(x) and g(x) are consistent and dependent. f(x) and h(x) are consistent and independent. g(x) and h(x) are consistent and independent.

21. Example 3 A. Graph the system of equations below. What type of system of equations is shown? x + y = 5 2x = y – 5 A.consistent and independent B.consistent and dependent C.consistent D.none of the above

22. Example 3 B. Graph the system of equations below. What type of system of equations is shown? x + y = 3 2x = –2y + 6 A.consistent and independent B.consistent and dependent C.inconsistent D.none of the above

23. Example 3 C. Graph the system of equations below. What type of system of equations is shown? y = 3x + 2 –6x + 2y = 10 A.consistent and independent B.consistent and dependent C.inconsistent D.none of the above

24. Example 3 A.f(x) and g(x) are consistent and dependent. B.f(x) and g(x) are inconsistent. C.f(x) and h(x) are consistent and independent. D.g(x) and h(x) are consistent. D. Graph the system of equations below. Which statement is not true? f(x) = x + 2 g(x) = x + 4

25. Concept

27. Example 4 FURNITURE Lancaster Woodworkers Furniture Store builds two types of wooden outdoor chairs. A rocking chair sells for $265 and an Adirondack chair with footstool sells for $320. The books show that last month, the business earned $13,930 for the 48 outdoor chairs sold. How many of each chair were sold? Use the Substitution Method Understand You are asked to find the number of each type of chair sold.

28. Example 4 Define variables and write the system of equations. Let x represent the number of rocking chairs sold and y represent the number of Adirondack chairs sold. Use the Substitution Method x + y =48The total number of chairs sold was 48. 265x + 320y =13,930The total amount earned was $13,930. Plan

29. Example 4 Solve one of the equations for one of the variables in terms of the other. Since the coefficient of x is 1, solve the first equation for x in terms of y. Use the Substitution Method x + y =48First equation x=48 – ySubtract y from each side.

30. Example 4 Solve Substitute 48 – y for x in the second equation. Use the Substitution Method 265x + 320y =13,930Second equation 265(48 – y) + 320y =13,930Substitute 48 – y for x. 12,720 – 265y + 320y=13,930Distributive Property 55y=1210Simplify. y=22Divide each side by 55.

31. Example 4 Now find the value of x. Substitute the value for y into either equation. Use the Substitution Method x + y =48First equation x + 22 =48Replace y with 22. x=26Subtract 22 from each side. Answer:They sold 26 rocking chairs and 22 Adirondack chairs.

32. Example 4 Use the Substitution Method Check You can use a graphing calculator to check this solution.

33. Example 4 A.210 adult; 120 children B.120 adult; 210 children C.300 children; 30 adult D.300 children; 30 adult AMUSEMENT PARKS At Amy’s Amusement Park, tickets sell for $24.50 for adults and $16.50 for children. On Sunday, the amusement park made $6405 from selling 330 tickets. How many of each kind of ticket was sold?

34. Concept

35. Example 5 Use the elimination method to solve the system of equations. x + 2y = 10 x + y = 6 Solve by Using Elimination In each equation, the coefficient of x is 1. If one equation is subtracted from the other, the variable x will be eliminated. x + 2y=10 (–)x + y= 6 y= 4Subtract the equations.

36. Example 5 Now find x by substituting 4 for y in either original equation. Solve by Using Elimination x + y=6Second equation x + 4=6Replace y with 4. x= 2Subtract 4 from each side. Answer:The solution is (2, 4).

37. Example 5 A.(2, –1) B.(17, –4) C.(2, 1) D.no solution Use the elimination method to solve the system of equations. What is the solution to the system? x + 3y = 5 x + 5y = –3

38. Example 6 Read the Test Item You are given a system of two linear equations and are asked to find the solution. No Solution and Infinite Solutions Solve the system of equations. 2x + 3y = 12 5x – 2y = 11 A. (2, 3) B. (6, 0) C. (0, 5.5) D. (3, 2)

39. Example 6 No Solution and Infinite Solutions x =3x =3 Solve the Test Item Multiply the first equation by 2 and the second equation by 3. Then add the equations to eliminate the y variable. 2x + 3y=124x + 6y=24 Multiply by 2. Multiply by 3. 5x – 2y=11(+)15x – 6y=33 19x =57

40. Example 6 Replace x with 3 and solve for y. No Solution and Infinite Solutions 2x + 3y=12First equation 2(3) + 3y=12Replace x with 3. 6 + 3y=12Multiply. 3y=6Subtract 6 from each side. y=2Divide each side by 3. Answer:The solution is (3, 2). The correct answer is D.

41. Example 6 Solve the system of equations. x + 3y = 7 2x + 5y = 10 A. B.(1, 2) C.(–5, 4) D.no solution

42. Concept

43. End of the Lesson Let’s Practice

44. Extra Practice

45. Example: Finding a Break-Even Point A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at $80 per pair. a. Write the cost function, C, of producing x pairs of running shoes. b. Write the revenue function, R, from the sale of x pairs of running shoes.

46. Example: Finding a Break-Even Point (continued) A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at $80 per pair. c. Determine the break-even point. Describe what this means. The break-even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system: or

47. Example: Finding a Break-Even Point A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at $80 per pair. c. Determine the break-even point. The break-even point is (6000, 480,000).

48. Example: Finding a Break-Even Point (continued) A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at $80 per pair. c. The break-even point is (6000,480,000). Describe what this means. This means that the company will break even if it produces and sells 6000 pairs of running shoes. At this level, the money coming in is equal to the money going out: $480,000.