Question

Plz help with this word problem

Answer 1

Well, this problem is best solved by setting up a system of two linear equations.

A linear equation can be defined 
y=mx+b
where 
b=initial value (when x=0), and
m=rate of increase or decrease.

In the given example, the x-axis represents hour, and the y-axis, number of cells.

Chemical #1 
initial value = b = 12000 cells
rate = m = -4000 / hr
The equation is therefore 
y1=-4000x+12000......................(1)

Similarly, for chemical #2
initial value = b = 6000 cells
rate = m = -3000 / hr
The equation is therefore
y2=-3000x+6000 .......................(2)

The time the two will have an equal sized colony would represent the solution of the system of equations (1) and (2), i.e. when  y1=y2
which means
-4000x+12000 = -3000x+6000
transpose and solve for x
4000x-3000x = 12000-6000
1000x=6000
x=6 hours.

At 6 hours from the start, 
y=-4000x+12000 = -4000*6+12000 = -24000+12000 = -12000 cells

So the solution is x=6, y=-12000, or (6,-12000)

Physical interpretation
Since cells cannot have a negative number, the two are actually equal before six hours, when they are both zero.
Case 1: y=0 when x=3
Case 2: y=0 when x=2

Therefore, after three hours, both trials will have zero cells.

You have to judge whether to give the mathematical solution (x=6,y=-12000) or the physical interpretation (x=3, y=0) as the answer.

Answer 2

The question asks for linear equations and the solution of them. Before you go to that trouble, you can solve this in your head. You start with 6000 more cells of Strain A than Strain B, and you kill them at a rate that is 1000 cells per hour faster. You will obviously kill off that 6000 cell difference after
.. (6000 cells)/(1000 cells/hour) = 6 hours

The linear equations are easily written in slope-intercept form since you are given starting values and rates of change.

a = 12000 -4000t . . . . . . . a is cells of Strain A, t is hours
b = 6000 -3000t . . . . . . . . b is cells of Strain B.

You want to know when a=b, so you can solve these by equating the expressions for a and b.
a = b
12000 -4000t = 6000 -3000t
(12000 -6000) = (4000 -3000)t . . . . . . add 4000t -6000
This should start to look familiar, as it matches the verbal description given above.
6000 = 1000 t
6000/1000 = t = 6

The number of cells will be the same after 6 hours.