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Writing and Solving equations with years
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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
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Eric Cartman High School had 1500 students in 2000 and 1600 in 2005
Eric Cartman High School had 1500 students in 2000 and 1600 in Assuming a linear increase, how many students will be in Cartman High in 2011? “Respect my authoriti and learn!”
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Eric Cartman High School had 1500 students in 2000 and 1600 in 2005
Eric Cartman High School had 1500 students in 2000 and 1600 in Assuming a how many students will be in Cartman High in 2011? “linear increase” means what? Slope! linear increase,
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Slope equation: m = What do we need to find the slope? Two sets of ordered pairs
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Eric Cartman High School had 1500 students in 2000 and 1600 in 2005
Eric Cartman High School had 1500 students in 2000 and 1600 in Assuming a linear increase, how many students will be in Cartman High in 2011? Years will always be x, so replace x1 with 2000 and x2 with 2005 x1 , y x2 , y2 (2000, y), (2005, y) (x, y), (x, y)
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Eric Cartman High School had 1500 students in 2000 and 1600 in 2005
Eric Cartman High School had 1500 students in 2000 and 1600 in Assuming a linear increase, how many students will be in Cartman High in 2011? When comparing years we can call the first year zero (0) and the next year 5 in this case x1 , y x2 , y2 (2000, y), (2005, y) (0, y), (5, y)
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Eric Cartman High School had 1500 students in 2000 and 1600 in 2005
Eric Cartman High School had 1500 students in 2000 and 1600 in Assuming a linear increase, how many students will be in Cartman High in 2011? Replace the y’s with the value goes with each year x1 , y x2 , y2 (0, 1500), (5, y) (0, 1500), (5, 1600) (0, y), (5, y)
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(0, 1500), (5, 1600) Replace the y’s in the equation with the numbers in the ordered pairs y2 – y1 m = x2 – x1
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(0, 1500), (5, 1600) Replace the x’s in the equation with the numbers in the ordered pairs m = x2 – x1 5 - 0
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100 1600 - 1500 m = 20 5 5 - 0 Simplify the top and the bottom
Simplify the fraction. If it’s not an even number, leave it as a fraction in it’s lowest form m = 20 5 5 - 0
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Once you know the slope, plug it into your equation:
Next we must find b. To find b we plug in either point for x and y y = 20x + b y = mx + b
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(0, 1500), (5, 1600) Plug in the first point Multiply on the right side 20(0) cancels out so we’re left with 1500 = b 1500 = b 1500 = 0 + b 1500 = 20x + b 1500 = 20(0) + b y = 20x + b
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y = 20(11) + 1500 y = 20x + 1500 y = 20x + b Plug b into the equation
The question then asks, “how many students will be in Cartman High in 2011?” 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 y = 20(11) y = 20x y = 20x + b
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y = 1720 y = 220 + 1500 y = 20(11) + 1500 Multiply 20(11)
Add together to get your answer In 2011 there should be 1720 students at Cartman High
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In 1990 there were 170,000 people in Kenny City
In 1990 there were 170,000 people in Kenny City. Since then, the population has been decreasing by 20,000 each year. Write a linear equation to represent how many people, p, are in Kenny City for any year, y.
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If this trend continues, how many people will be left in Kenny City in 2015?
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Stan Marsh Ski Slope had 6000 skiers for the season in 1980
Stan Marsh Ski Slope had 6000 skiers for the season in 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?
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Broflovski’s law firm handled 524 cases in 1995
Broflovski’s law firm handled 524 cases in In 2003 they handled 628 cases. Assuming a linear increase, how many cases should they have handled in 2010?
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In 1980, the average apartment at Butters Apartments was $250
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.
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