Let's start by assigning variables to represent the unknowns in the problem.
Let's use "c" to represent the number of crayons Manuel had in the beginning and "m" to represent the number of markers he had in the beginning.
From the problem, we know that:
Manuel had 4 times as many crayons as markers in the beginning, so:
c = 4m
After he bought 250 crayons and 100 markers, he had 3 times as many crayons as markers, so:
c + 250 = 3(m + 100)
Now we can use algebra to solve for c:
c + 250 = 3m + 300 // distribute the 3
c = 3m + 300 - 250 // simplify by combining like terms
c = 3m + 50
Substitute c = 4m from the first equation into the second equation:
4m = 3m + 50 // subtract 3m from both sides
m = 50
So Manuel had 4 times as many crayons as markers in the beginning, which means he had:
c = 4m = 4(50) = 200 crayons in the beginning.