Intergrated Rate Law for Chemistry

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line, y=mx+b. Please help me out

a) The reactant concentration in a first-order reaction was 7.00×10−2 M after 30.0 s and 5.30×10−3 M after 90.0 s. What is the rate constant for this reaction?

b) The reactant concentration in a second-order reaction was 0.710 M after 225 s and 3.30×10−2 M after 770 s. What is the rate constant for this reaction?

ok.. let's start here...

for the reaction
A --> B + C

the rate is equal to the change in concentration of A per unit time..
in chemistry [ ] means concentration
and in math dX means change in X

so..
rate = change in concentration of A per unit time can be written...
rate = - d[A] / dt

and the - sign is because [A] is decreasing

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also...
the rate is proportational to the concentration of A. The higher the concentration, the faster the rate
so..
rate α [A]

and we solve proporationalities in my by inserting a constant.. "k" in this case
rate = k x [A]

and some reactions have different mechanisms. Sometimes 2 molecules of A must collide to react. sometimes only 1 can decompose all by itself.. so there is an exponent on that [A]

ie...
rate = k x [A]^n

********
putting both those equations together...
rate = - d[A] / dt = k x [A]^n

now..
if n=0.. that's a zero order rxn
if n=1.. first order
if n=2.. 2nd order

so let's solve those differential equations

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0 order
-d[A]/dt = k x [A]^0
-d[A]/dt = k
d[A] = -k x dt
and integrating from the initial concentration of [Ao] at time = 0 to the final concentration of [At] at time = t
∫d[A] = -k ∫dt
becomes...
.. ...[At].. ... .. ..t
[A].|.. = .-k x t |
.. ...[Ao].. .. .. ..0
ie...
[At] - [Ao] = -kt