Question 182562
WE KNOW RIGHT OFF THE BAT THAT WE HAVE THREE UNKNOWNS AND ONLY TWO EQUATIONS SO MORE THAN ONE ANSWER IS A DISTINCT POSSIBILITY
Let x=number of 0.5 dollar stamps
y=number of 3 dollar stamps
z=number of 10 dollar stamps

x+y+z=100-----------------------eq1
0.5x+3y+10z=100----------------eq2

multiply eq2 by 2 and subtract eq 1 from it:
(x+6y+20z=200)-(x+y+z=100) and we get
5y+19z=100 solve for y
5y=100-19z
y=(100-19z)/5 or
y=20-19z/5----------------------eq2a
Now we know two important aspects of this problem:
(1)  We cannot have fractions of dollars other than the half dollars
(2)  We cannot have negative dollars
So, from eq2a, 20-19z/5 >=0
-19z/5>=-20
-19z>=-100
19z<=100
z<=5.263----round to the next smallest whole number
z<=5
Now we will start assigning values to z between 0 & 5 and see which values produce whole numbers for y in eq2a.
z=0  y=20--------------------possibility
z=1  y=fraction--------------------no good
z=2  y=fraction ---------------------no good
z=3  y=fraction-----------------------no good
z=4  y=fraction------------------------no good
z=5  y=1------------------------possibility

Only two possibilities:
z=0, y=20 or z=5, y=1
Now we plug these possibilities in eq1 and see what we get for x
y=0, z=20
x+20+0=100
x=80
So now we have x=80, y=20 & z=0
Let's see if these work in eq 2:
0.5*80+3*20+10*0=100
40+60=100
100=100
So, one solution is:
x=80, y=20 & z=0----------------------------------One solution
Now for y=1, z=5
Plug these into eq1 and we get:
x+1+5=100
x=94
Plugging these into eq2:
0.5*94+3*1+5*10=100
47+3+50=100
100=100
So x=94, y=1, z=5-------------------------------Another solution

Hope this helps-------ptaylor