In general with 2 equations and 3 variables you can't find a unique solution, instead you'll usually be faced with infinite solutions which can be written as a function of one variable. However, there is a slight twist in your case that you are told that [tex]x,y,z[/tex] are positive integers, this gives us extra information which can narrow down the number of solutions.
Start as you would normally, by solving for [tex]x,y,z[/tex] in terms of one of the variables (choose your favourite, it shouldn't matter), say [tex]x[/tex]. So we solve assuming [tex]x[/tex] is a fixed known constant, so now we have 2 unknowns [tex]y,z[/tex] and 2 equations.
[tex]x+y+z=100 \implies y = 100-x-z[/tex]
Substitute this into
[tex]5x+2y+z/4 = 100 \implies 5x+2(100-x-z)+z/4=100 \implies z = (12x+400)/7[/tex]
Substitute this back into [tex]y[/tex]
[tex]y=100-x-z \implies y=100-x-(12x+400)/7 \implies y=(300-19x)/7[/tex]
So [tex]y=(300-19x)/7[/tex] and [tex]z = (12x+400)/7[/tex], remember that the question tells us that [tex]x,y,z>0[/tex], this means that
[tex]y=(300-19x)/7>0 \implies 300-19x>0 \implies 300>19x \implies 300/19=15.789...>x[/tex]
Since [tex]x[/tex] is positive and an integer we've narrowed its down to 15 possibilities, [tex]x=1,2,3,...,15[/tex].
We could just try all 15 possibilities and see which ones work. However, if you know a bit of modulo arithmetic you can save yourself a bit of time.
So so far we've used the fact that [tex]x,y[/tex] are positive, if we try to do the same trick with [tex]z>0[/tex], we don't get anything useful. We get [tex]z>0[/tex] as a consequence of [tex]x>0[/tex], so it doesn't help us. The other information we are given is that [tex]x,y,z[/tex] are integers. In particular this means that [tex]y = (300-19x)/7[/tex] must be an integer, or put another way [tex]300-19x[/tex] must be divisible by [tex]7[/tex]. Using modula arithmetic this would be represented by [tex]300-19x\equiv 0\bmod 7[/tex], you can look up tutorials on the internet on how to solve this, but it will essentially tell you that if [tex]300-19x[/tex] is divisible by 7 then so is 4 times that, i.e.
[tex]4\times (300-19x) = 1200-76x = 7\times(172-11x)+(x-4)[/tex]
If the whole thing is divisible by 7 and [tex]7\times(172-11x)[/tex] is clearly divisible by 7, then it follows that [tex]x-4[/tex] is divisible by 7, i.e. [tex]x[/tex] has remainder 4 upon dividing by 7. This narrows down [tex]x[/tex] to 4, or 11.
We could try the same trick on [tex]z[/tex] being an integer but it turns out it will just tell us again that [tex]x[/tex] has remainder 4 when dividing by 7, so it isn't useful.
So there are two possible solutions [tex]x=4, y=32, z=64[/tex], and [tex]x=11, y=13, z=76[/tex], which you can check both satisfy all the constraints in the question.
Hope this helped,