Disguised inequality question

Can someone please explain this solution? It make no sense to me whatsoever. Greg did not have time in the video to explain or take any questions. Thanks!

Since x and y are positive numbers and x + y = 1, then we can see that x and y must be between 0 and 1. If either were greater than 1, then x+y > 1 (plug in numbers to see).

Thus, we have that 0<x<1 and 0<y<1.

We see that we are trying to find the possible values of 100x+ 200y. Let’s try to multiply the inequality 0<y<1 and the equation x + y = 1 by 100 on both sides.

i) 0<100y<100

ii) 100x+100y=100

We want 100x + 200y. We see that 100x + 200y = 100x + 100y + 100y = (100x + 100y) + 100y = 100 + 100 y.

I used (ii) from above to make the last substitution.

Thus, we see that 100x + 200y = 100 + 100y. We can bound 100 + 100y from below and above using 0<100y<100. Since 0<100y<100, then by adding 100 to each part of the inequality, we get 0 + 100 < 100 y + 100 < 100 + 100 → 100 < 100 + 100y< 200.

Thus, 100 + 100y is between 100 and 200 exclusive. Answers (II) and (III) work.