My opinion is that you should not try to eliminate 
at all. Instead, you should think about how 
and 
behave as 
varies, find some important points such as the critical points with respect to 
and 
, and try to sketch the graph from that information.
means that as 
goes from 
to 
, 
decreases from 
to a local minimum of 
at 
, then rises to 
at 
, then falls again to 
.
means that 
starts and ends at 
, with local maxima of 
when 
, and a local minimum of 
at 
.
Interestingly, 
is a critical point for both 
and 
, both of whose derivatives change sign, so the curve forms a cusp at that point.
That's enough information to follow the curve as 
goes from 
to 
: it comes in from the bottom right quadrant 
, goes upward and leftward till it hits the cusp 
, upon which it turns around, passing smoothly through 
where its tangent is horizontal, until it reaches the second cusp 
, and then turns around a second time and exits at the bottom left towards 
.
This is consistent with the plot arising from Juan Joder's answer.
Answer from user856 on Stack ExchangeVideos
What are the solutions to the equation 4t(3-t)=4t+3 ?
Find the zeros of 4t(3-t)=4t+3
My opinion is that you should not try to eliminate at all. Instead, you should think about how and behave as varies, find some important points such as the critical points with respect to and , and try to sketch the graph from that information.
means that as goes from to , decreases from to a local minimum of at , then rises to at , then falls again to .
means that starts and ends at , with local maxima of when , and a local minimum of at .
Interestingly, is a critical point for both and , both of whose derivatives change sign, so the curve forms a cusp at that point.
That's enough information to follow the curve as goes from to : it comes in from the bottom right quadrant , goes upward and leftward till it hits the cusp , upon which it turns around, passing smoothly through where its tangent is horizontal, until it reaches the second cusp , and then turns around a second time and exits at the bottom left towards .
This is consistent with the plot arising from Juan Joder's answer.
I wouldn't know how to do it in an exam, but using a Gröbner basis routine such as that in Mathematica easily gives the implicit Cartesian equation
There's roughly $kT$ energy in each active mode. The active modes are characterized by momenta which live inside a sphere of radius proportional to $kT$, which has volume proportional to $T^3$. Multiplying these factors gives $T^4$, and the result clearly generalizes to $T^{d+1}$ in general dimension.
If you know Quantum Mechanics, you know that you can set length to have dimensions of the inverse of energy. This means that $j$ must have dimensions of energy to the four. If you consider that the only variable with energy units is the temperature, then the energy density must be proportional to $T^4$.
If you consider additional constants with energy dimensions, like a mass, then the derivation is no longer valid.