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 \geq 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 $y \geq 0 .$

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 - 2 t

Substitute

$y = 0$

0 = 6 - 2 t

AddEqn

$LHS + 2 t = RHS + 2 t$

2 t = 6

DivEqn

$LHS / 2 = RHS / 2$

t = 3

Technically, after

3

hours the number of bacteria remaining is the same.

Exercise

This Solution is Missing. Mathvizy Can Solve This Problem For You