Answers

2014-11-03T15:19:10+08:00
Let 'x' be the number of books on the first pile
     'y' be the number of books on the second pile
     'z' be the number of books on the third pile
x + y + z = 3000    -----equation 1
x = y + 10   -----equation 2
y = 2z     -----equation 3
from equation 3
y = 2z
z = y/2   -----equation 3'
substitute equations 3' and 2 to equation 1
x + y + z =3000
(y+10) + y + (y/2) = 3000
y + 10 + y + y/2 = 3000
2y + y/2 = 3000-10
(4y+y)/2 = 2990
5y = 2990(2)
5y = 5980
y=1196
substitute y=1196 to equation 2
x = y + 10
x = 1196 +10
x = 1206
substitute y=1196 to equation 3'
z = y/2
z = 1196/2
z = 598
therefore there are 598 books in the third pile.
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