Why are there two different and incompatible formulae for Hooke's Law? I understand Hooke's Law to be 
F = kX

Where F is the tension applied to an object, X is the extension/change in length, and k is the spring constant, whose units are N/m.
I'm learning Mechanics 3 under the A Level Edexcel Maths specification (the M3 stuff is on page 70), and I am being told that Hooke's Law can also be defined as:
F = (λx)/l

Where x is the change in length, F is the tension and λ is the Modulus of Elasticity/ the Young's Modulus.
However, this would imply that the spring constant is equal to the Modulus of Elasticity, divided by the length. Instantly, if one already knows the units for the Modulus of Elasticity (Nm^-2), we can see that that would imply the spring constant is equal to Nm^-3, while the formula F = kX implies its units is N/metre.
I investigated further, and we know that modulus of elasticity is stress/strain, which is 
(Fl)/(Ax)

With a new variable being introduced as A, which is the cross-sectional area of the object.
We know that k = F/X, and do we can express the modulus of elasticity as
(kl)/A

Which would therefore mean that the spring constant k is 
(λA)/l
and not (λ/l)
We can check the units too, and (λA)/l gives N/metre.
So my question is: Why am I seeing two incompatible versions of Hooke's Law? The spring constant cannot be equal to F/x and λ/l, as the units aren't the same for one, and my above calculations also show that we need to include the cross-sectional area.
However, this is in the M3 syllabus and it's not anything new (although I haven't actually seen the formula anywhere else for k being λ/l) , so it's likely that there's something I'm not understanding.
Could anyone shed any light?
 A: $F=kx$ is used in physics. $F=(\lambda / l)x$ is used in applied maths. Clearly $k=\lambda / l$.
$\lambda$ is called the modulus (of elasticity) but it is not the same as Young's Modulus.
To avoid confusion, decide which subject you are studying (physics or applied maths) then use the appropriate formula and ignore the other one. 
A: I'd say that the inclusion of the cross sectional area is a must. Perhaps the equation used a modulus which was already for a certain cross section? Otherwise it is easy to see that the cross sectional area must come into the expression: If you take a second spring, you double the cross sectional area. Then the force should also increase by the same factor. In the $F = kx$ version this is done by just setting $k \to 2k$. In the other example, the properties of the material does not change but there must be a factor 2 coming in as well.
Analysis with units must always succeed. If the units don't match up, something is wrong.
A: Well. To focus on the original post, both constants can be used in maths or physics. If you are stuck to a particular scheme then they may be exclusively used but in general they are just alternative ways of looking at spring extensions.
The restoring Force
$$
F=-kx \text{   or  } F=-\lambda x/l
$$
The second of these divides out the length and so any length of particular spring will have the same value whereas k will be proportional to the length.
Then the energy stored is integrated out as
$$
F=kx^2/2 \text{   or  } F=\lambda x^2/2l
$$
The units of $$\lambda  \text{  are Newtons whereas the units of  } k \text{ are  Newtons/metre}$$
It perhaps inconvenient that these are both used from time to time but there you have it.
