Answers to 2 Problems from Week 4 To Do list
1. College A has a y intercept (time = 0, or now..) or 5000, the current enrollment. The slope or change is 50 students per quarter. The equation, then is y = 50x + 5000
College B has a y intercept of 3500 and a slope of 200. y = 200x + 3500
substitute 50x + 5000 in for the y in the second equation:
50x + 5000 = 200x + 3500 Solve for x
1500 = 150x x = 10 Therefore it will take 10 quarters before college B will be the same size as college A. That enrollment will be 5500.
2. The sum of the 3 numbers is 11: x + y + z = 11. The third number is equal to twice the second number plus the first number: z = 2y + x. Twice the first number plus the third number is equal to three times the second number: 2x + z = 3y. The 3 equations, then, are:
x + y + z = 11
z = 2y + x
2x + z = 3y
Since the second equation is already solved for z, it is easiest to use this equation and plug it into the other 2: These equations are:
x + y + (2y + x) = 11
2x + (2y + x ) = 3y
Now you have 2 equations and 2 variables.. Now it is just like the problem above. Solve for x or y in one of the equations above and substitute the answer into the other equation: One way is to solve for x in the first equation, so x + y + 2y + x = 11, or 2x + 3y = 11. 2x = 11 – 3y
x = (11 -3y)/2 substitute in for the x in the second equation:
Lets simplify the 2nd equation first: 2x + (2y + x ) = 3y or 3x + 2y = 3y or 3x = y. Now substitute in what we solved for above:
3((11-3y)/2) = y (33 – 9y)/2 = y get the y’s on one side.. multiply both sides by 2 33 – 9y = 2y. 33 = 11y. y = 3. So.. the second number is 3. Now plug 3 in for y in one of the equations above to solve for x:
3x = y is one equation. let y = 3. 3x = 3, so.. x = 1
x = 1 and y = 3. plug in for one of the first equations: the first is the easiest
x + y + z = 11, 1 + 3 + z = 11 z = 7
So the 3 numbers are: x=1, y=3, and z = 7