Very nice, happy7 !!!!!
Here's another way to solve this....we have more variables than equations, so we have infinite solutions....but only one makes sense... (as we shall see !!!)
Call the number of oxen, x, the number of pigs, y, and the number of sheep, z.
And we know that the total number of animals must = 100...so we have.....
x + y + z = 100 ...and solving for z, we have
z = 100 - x - y (1)
And we know that each animal's value times their number summed together must = 100......so......
$10x + $3y + $.50z = 100
And replacing z with (1), we have
10x + 3y + .50(100 - x - y) = 100 ......simplify.......
10x + 3y + 50 - .50x - .50y = 100 .......simplify some more.....
9.5x + 2.5y = 50 ........ multiply by 2 on each side......
19x + 5y = 100
Now, notice that x must be some integer that is divisible by 5, because if we subtracted 19x from 100, this quantity would then be divisible by 5......i.e, making y an integer value, which is what we want..
And notice that x couldn't be any multiple of 5 larger than 5, because that would make y negative.....and we can't have a negative number of pigs!!!
So....x must be 5......so we have....
19(5) + 5y = 100
95 + 5y = 100 subtract 95 from both sides
5y = 5 divide both sides by 5
y = 1
And z = 100 - x - y .......so....... z = 100 - 5 - 1 = 100 - 6 = 94
And just as Happy7 has said......we have 94 sheep, 1 p*g and 5 oxen.....
