Page 269 - Buku Siap OSN Matematika SMP 2015(1)
P. 269
Solusi Olimpiade Matematika 2013
Jawab:
Kelereng awal:
A = x, B = y, C = z
Hari pertama:
A = x – y – z
B = 2y
C = 2z
Hari kedua:
A = 2(x – y – z) = 2x – 2y – 2z
B = 2y – (x – y – z) – 2z
= 2y – x + y + z – 2z
= 3y – x – z
C = 4z
Hari ketiga:
A = 2(2x – 2y – 2z)
= 4x – 4y – 4z
B = 2(3y – x – z)
= 6y – 2x – 2z
= –2x + 6y – 2z
C = 4z – (2x – 2y – 2z) – (3y – x – z)
= 4z – 2x + 2y + 2z – 3y + x + z
= 7z – x – y
= –x – y + 7z
A 4x – 4y – 4z = 16
x – y – z = 4 ... persamaan (1)
B –2x + 6y – 2z = 16
260 Wahyu
