An elastic spring under tension of 3 N has a length a. Its length is b under tension 2 N. For its length (3a-2b), the value of tension will be ____ N.

Idea: tension is linear in length, so any length written as a combination of a and b carries the same combination of their tensions — provided the coefficients add to one, as 3-2 does.

Doing it explicitly, with T=k(L-L₀):

3=k(a-L₀) and 2=k(b-L₀).

Subtracting, 1=k(a-b), so k=1/(a-b).

From the first, a-L₀=3(a-b), giving

L₀=a-3(a-b)=3b-2a.

Extension at the length 3a-2b:

(3a-2b)-(3b-2a)=5a-5b=5(a-b).

T=k×5(a-b)=(5(a-b))/(a-b)=5 N.

The shortcut: T(3a-2b)=3T(a)-2T(b)=3(3)-2(2)=5 N, which works because T is an affine function of L and the weights 3 and -2 sum to 1.