SOLUTION: i have 100 dollars and need to buy a hundred items and one item cost 3 dollars the other cost .50 cents and the last one cost 10 dollars how many of each will i have

Let x = number of $3 items
Let y = number of $.50 items
Let z = number of $10 items

 x +    y +   z = 100    <--item equation
3x +  .5y + 10z = 100    <--money equation

Clear the decimal of the second equation by multiplying it through by 2

 x +  y +   z = 100
6x +  y + 20z = 100

Subtracting the first equation from the second gives

5x      + 19z = 100

Solve for x:

          5x = 100 - 19z
           x =  20 - 

z has to be a multiple of 5 for  to be a whole number
and if z were bigger than 5, x would be negative, so the only value
z can take on is 5 which makes x = 20 - 19 = 1, so  substituting in

x + y + z = 100
1 + y + 5 = 100
    y + 6 = 100
        y = 94

1 of the first item, 94 of the second item and 5 of the 3rd.

Edwin