Is it? Because I’m a transwoman and my levels are all within cis woman ranges for going on a decade now, yet I’m basically full-time Chandra, Awakened Inferno unless it’s subzero temps outside.

Ye cannot get ye flask.

Hey, I can take a swing at this. It’s basically just a question of understanding how fractions work (which is fumbled horrendously by teachers, at least where I’m from - I basically had to teach myself fractions all over again when I went back to school).

So, if you look at the terms on the left hand side, we have “x”, which is the same as saying “1x”, so the whole number “1”, we have a whole number “3” as part of “3x”, and we have the fraction that’s going to cause us to do a little work, “1/2” as part of “1/2x”.

Now, a whole number can be rewritten as a fraction, and this makes the most sense when you see fractions as little division problems unto themselves. For instance, the “1/2” could be read as “1 divided by 2”, or “0.5”. A whole number like “1”, then, could be rewritten as “1/1”, or “2/2”, or “3/3”, and so on.

Now, in order to add fractions together (which is what we’re trying to do since our ultimate goal is to get the variable that we’re solving for alone on one side of the equation), we need the denominator to be the same for all of our terms, i.e. the “common denominator”. Because we already know the denominator we likely need, the “2” in “1/2”, we simply need to transform both of our whole numbers into fractions with 2 in the denominator.

For “1”, this can be rewritten as “2/2”. Dividing 2 by 2 gets us back to 1, so that works out.

For “3”, we need to determine what number divided by 2 gets us to 3. In this case, that’s 6, which leaves us with “6/2”.

The equation now looks like this: 2/2x + 6/2x + 1/2x = 45

We can, of course, pull the “x” out like this: x(2/2 + 6/2 + 1/2) = 45

Then, when adding fractions, we only add the numerators (the reason we were looking for the common denominator in the first place). So, 2 + 6 + 1 = 9, leaving us with “9/2x = 45”. It’s then just a question, as you can see in the posted solution, of multiplying both sides by the reciprocal to solve for x.