There are 4 more women than men on Centerville's board of

I didn't set up this equation correctly. How is the algebra reasoning for this question diff than Q 29? When do we know that we need to set up 2 diff equations representing 2 diff variables?

Thanks!

W = M + 4 (Equation 1)
M + W = 10 because there are ten members on the board. (Equation 2)

Simplifying Equation 2 --> M = 10 - W
Substitute, M = 10 - W into Equation 1.
Hence, W = 10 - W + 4
2W = 14
W = 7 (Option D)

Let M be the no of men.
Let Women be expressed in terms of M ie W = M+4
We also know Men + Women = 10 therefore M + M+4 = 10 Solving the equation we get 2m = 6 and m = 3. Since there are 4 more women than men the total no of women = 3+ 4 = 7 .
The question asks for the no of women = 7 is the answer
The ans also fits into the constrain in the q = total = 10 ( 3 + 7 )

We can also solve this by testing the ans choices
Eg : With C = 6 women + 2 men =8 is incorrect as the total no of members needs to be 10 on the board.
It also means ans choice a & B are also wrong as it wont suffice the constrain.
Lets check ans choice d = 7 women = 3 men = 7+ 3 = 10 = correct . Therfore ans choce = D