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:
-8a(a+3b)
Step-by-step explanation:
(2a+6b)(6b−2a)−(2a+6b)^2
(2a+6b) {(6b−2a)−(2a+6b)(2a+6b)}
2(a+3b) (6b-2a-2a-6b)
2(a+3b) (-4a)
-2(a+3b) x 4a
-2 x 4a (a+3b)
-8a(a+3b)
