2.  Lesson Objective: 4.01a Use linear functions or inequalities to model and solve problems; justify results  Students will know how to use base year analysis to write and solve word problems with years

3. WWhen comparing information from two years, we can use base-year analysis WWe make the first year of information year zero TTo get the second year we subtract the second year from the first year

4. YYear 1: (1995, 25) YYear 2: (2010, 55) BBase year: year 1 = 0 so we have (0, 25) YYear 2 = 2010 – 1995 = 15, so: (15, 55) WWe can than use the two points to get a slope and an equation

5. ((0, 25), (15,55) SSlope equation: m =  Replace y 2 with 55 and y 1 with 25 RReplace x 2 with 15 and x 1 with 0 SSimplify the top and the bottom RReduce the fraction WWe have a slope of 2

6. OOnce you know the slope, plug it into the equation RReplace the y and x with one of the points. II would suggest the first point since x = 0 MMultiply and solve for b PPlug b back into the equation

7.  Eric Cartman High School had 1500 students in 2000 and 1600 in 2005. Assuming a linear increase, how many students will be in Cartman High in 2011? “Respect my authoriti and learn!”

8. EEric Cartman High School had 1500 students in 2000 and 1600 in 2005. Assuming a how many students will be in Cartman High in 2011? ““linear increase” means what? SSlope! linear increase,

9. SSlope equation: m = WWhat do we need to find the slope? TTwo sets of ordered pairs

10. EEric Cartman High School had 1500 students in 2000 and 1600 in 2005. Assuming a linear increase, how many students will be in Cartman High in 2011? YYears will always be x, so replace x 1 with 2000 and x 2 with 2005 x 1, y 1 x 2, y 2

11. EEric Cartman High School had 1500 students in 2000 and 1600 in 2005. Assuming a linear increase, how many students will be in Cartman High in 2011? WWhen comparing years we can call the first year zero (0) and the next year 5 in this case

12. EEric Cartman High School had 1500 students in 2000 and 1600 in 2005. Assuming a linear increase, how many students will be in Cartman High in 2011? RReplace the y’s with the value goes with each year

13. RReplace the y’s in the equation with the numbers in the ordered pairs

14. RReplace the x’s in the equation with the numbers in the ordered pairs

15. SSimplify the top and the bottom SSimplify the fraction. If it’s not an even number, leave it as a fraction in it’s lowest form

16. OOnce you know the slope, plug it into your equation: NNext we must find b. TTo find b we plug in either point for x and y

17. PPlug in the first point MMultiply on the right side 220(0) cancels out so we’re left with 1500 = b

18. PPlug b into the equation TThe question then asks, “how many students will be in Cartman High in 2011?” XX is always years, so we’ll plug in the year for x. Remember, we have to plug in how many years it’s been since 2000, so plug in 11

19. MMultiply 20(11) AAdd together to get your answer IIn 2011 there should be 1720 students at Cartman High

20.  In 1990 there were 170,000 people in Kenny City. Since then, the population has been decreasing by 2,000 each year. Write a linear equation to represent how many people are in Kenny City for any year.

21. IIn 1990 there were 170,000 people in Kenny City. Since then, the population has been decreasing by 2,000 each year. Write a linear equation to represent how many people are in Kenny City for any year. LLinear equation means y = mx + b FFirst we need to find the m

22. IIn 1990 there were 170,000 people in Kenny City. Since then, the population has been each year. Write a linear equation to represent how are in Kenny City for any year. YYears are always x, so what goes with years in the problem? many people decreasing by 2,000

23. IIn 1990 there were 170,000 people in Kenny City. Since then, the population has been decreasing by 2,000 each year. Write a linear equation to represent how many people are in Kenny City for any year. DDecreasing means subtracting, so replace m with -2,000

24. IIn 1990 there were 170,000 people in Kenny City. Since then, the population has been decreasing by 2,000 each year. Write a linear equation to represent how many people are in Kenny City for any year. NNext we need to find the b WWe plug in a point to solve for b

25. IIn 1990 there were 170,000 people in Kenny City. Since then, the population has been decreasing by 2,000 each year. Write a linear equation to represent how many people are in Kenny City for any year. WWhen we deal with years, the first year is called year zero, so 1990 is year 0

26. IIn 1990 there were 170,000 people in Kenny City. Since then, the population has been decreasing by 2,000 each year. Write a linear equation to represent how many people are in Kenny City for any year. RReplace y with 170,000 and x with 0

27. MMultiply -2000(0) 1170000 = b PPlug b back into the equation

28.  In 1990 there were 170,000 people in Kenny City. Since then, the population has been decreasing by 2,000 each year. Write a linear equation to represent how many people are in Kenny City for any year.

29. IIf this trend continues, how many people will be left in Kenny City in 2015? UUse the equation to plug in the new year RRemember, 1990 was year 0, so 2015 would be year 25

30. MMultiply -2000(25) AAdd the numbers together yy = 120000, so in 2015 there will be 120,000 people left in Kenny City

31.  Stan Marsh Ski Slope had 6000 skiers for the season in 1980. In 2000 it had 9600 skiers. If Stan Marsh Ski Slope continues to increase at the same rate, how many skiers will there be in 2011?

32.  Broflovski’s law firm handled 524 cases in 1995. In 2003 they handled 628 cases. Assuming a linear increase, how many cases should they have handled in 2010?

33.  In 1980, the average apartment at Butters Apartments was $250. By 2004, the average price was $702. (Let x = 80 represent 1980) Create a linear model that best represents this situation then find the average price of an apartment in 2010.