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