Answer:
y=100t+1200, where t is years since 2013 and
Explanation:
We only have two points from which to form an equation. Since we are told that it is a linear equation, two points is all we need to find an equation in the form of y=mx+b, where m is the slope (the rate of change) and b is the y-intercept (the value of y when x=0). Let's make the x axis the year and y the student population.
Let's put the data into an x,y format, but before we do, lets be sure to adopt the problem's fervent wish that we treat the x variable, t, as the years since 2013. "t is the number of years after 2013." (0,1200) becomes is the data point for 2013. The school's population will be the y-axis.
(0,1200) [primary school had 1200 students enrolled in 2013], and
(3,1500) [1500 students in 2016]
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First calculate m, the slope. Slope is the Rise/Run between the two points (the change in y over the change in x)
Going from (2013,1200) to (2016,1500) we find that:
Rise (in y) = (1500 - 1200) = 300 students
Run (on x) = (2016 - 2013) = 3 years
The Rise/Run, or slope, m, is 300/3 or 100: m = 100
Lets put that into our equation format:
y = 100x + b
X will be the time, in years. The problem asks that we define x in terms of years after 2013, and that this be assigned the variable, t:
y = 100t +b, where t is years after 2013.
(The 100 tells us that the student population increases by 100 per year)
We now need b, the y-intercept. One can either graph this and look for the value of y when x=0, of simply input either of the data points. This looks easy enough to input a point and solve for b:
Remember that t is the (year - 2013)
y = 100t +b Solve for (2013,1200)
t = 2013 - 2013 = 0
1200 = 100*(0) + b
b = 1200
Actually, b was easy enough to find without either calculating or graphing, since one of our two points tells us the y-intercept (0,1200).
The equation becomes y = 100t + 1200
See the attached graph.