Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Applications of Linear Systems
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Exercise 10 Page 390

See solution.

Practice makes perfect
Let y be the number of cells, in thousands, and t the time in hours. In Strain A there were initially 6 thousand cells and each hour the number decreased by 2 thousand cells. Let's write an equation that describes the situation. y=6-2 t In Strain B we started with 2 thousand cells and the rate of decrease is 1 thousand cells per hour. We can also write this as an equation. y=2-1 t

By combining the equations we get a system of equations. y=6-2 t y=2-1 t Let's solve this system graphically. We limit the domain to t≥ 0 since that is when the study started.

As we can see, the graphs intersect at (4,- 2). This is counter-intuitive as the graphs describe the number of bacteria and that cannot be less than 0. Therefore, we have to limit our range to y≥ 0.

When we limit the range to non-negative values the system of equations has no solutions.

However...

The strains do have the same number of cells when they are both 0 — after all, the bacteria have died. Let's view this graphically.

When all the cells in Strain A have died, y=0, both strains have the same number of cells. Let's find when this happens.
y=6-2t
0=6-2t
2t=6
t=3
Technically, after 3 hours the number of bacteria remaining is the same.