Master Algebra 1 Word Problems with our Comprehensive Practice

rate of change proportional to the diffrence between itself and a fixed value $\frac{dy}{dx}=k(y-<!--\; -\; -->a)$
So I have a question regarding exponential growth/decay problem formulated in my book.
In my book they say:
"Sometimes an exponential growth or decay problem will involve a quantity that changes at a rate proportional to the difference between itself and a fixed point:
$\frac{dy}{dx}=k(y-<!--\; -\; -->a)$
In this case, the change of dependent variable u(t)=y(t)−a should be used to convert the differential equation to the standard form. Observe that u(t) changes at the same rate as $y(t)(ie,\frac{du}{dy}=\frac{dy}{dt}) so it satisfies
$\frac{du}{dt}=ku$
I am having a hard time grasping what they mean. And I would hope for somebody to expand on this notion.
More specifically how does $\frac{dy}{dx}=k(y-<!--\; -\; -->a)$ relate to $u(t)=y(t)-<!--\; -\; -->a$?

3 teaspoons = 1 tablespoon
4 tablespoons = 1/4 cup
According to the information above, how many teaspoons are equal to 3/8 cup?
A. 4
B. 3
C. 12
D. 18
E. 9

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$\{\begin{array}{l}-<!--\; -\; -->2b+3c=1\\ 3a+6b-<!--\; -\; -->3c=-<!--\; -\; -->2\\ 6a+6b+3c=5\end{array}$
An inconsistent linear system with the third row being 0 0 0 6?

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