Variation of Hooke's Law Confirmation

Question

If we have a spring with natural length $l$, modulus of elasticity $\lambda$ and it has a weight of $m_1g$ attached to it, then its extension, $x_1$, is $$\frac{m_1gl}{\lambda}$$ If we add on another mass with weight $m_2g$ then we find the change in extension, $x_2$, is $$\frac{m_2gl}{\lambda}$$ after some manipulation.

Hooke's law states that the tension in a spring, $T$, is equal to the spring constant, $k$, multiplied by the extension of the spring, $x$:

$$T=kx=\frac{\lambda x}{l}~~(\text{as the spring constant is equal to $\frac{\lambda }{l}$})$$

Therefore, is it correct to state Hooke's law somewhat differently:

$$\Delta T=\frac{\lambda\Delta x}{l}$$ where $\Delta T$ is the change in tension in the spring.

Mathematically this seems sound; I'd just like to verify this is true.

EDIT:

There seems to be some confusion about the measurements and terms that I have used. To clarify:

The spring constant, $k$, is equal to $\lambda/l$; so my question is equivalently: is the following variation of Hooke's Law correct?

$$\Delta T=k\Delta x$$ where $\Delta T$ is the change in tension in the spring and $\Delta x$ is the change in extension.

Answer 1

There is a dimensional mistake, if you write $x_1=\frac{m_1gl}{\lambda}$ you are wrting dimensionally this: $$m=\frac{kg\cdot m\cdot m}{\frac{N}{m}\cdot s^2}$$ you are saying that $m=m^2$ but it's impossbile. Remember that Hooke's law is the following: $$\vec{F}=\lambda(\vec{x}-\vec{x}_0)$$ The force is directly proportional to the displacement.
So if you add a weight you find that $x_1=\frac{m_1g}{\lambda}$ if you add another mass $m_2$ you have the following equation: $$\vec{x}_2=\frac{(m_1+m_2)\vec{g}}{\lambda}$$