(11 system of linear equations college production play of 360 tickets were sold the ticket prices were $8 the total income $ 10, 512 was 8 and 10 combined a) C) x+4 : 3320 8 x + 104 + 12 z = 360 X + y + 52 = 0 y + y + 2 = 360 3x + 10y +122=3320 x + y + 5z = d there were in 3 variables to solve a was > How many tickets were sold it $3.320 5 times the number of $12 tickets sold? b) x+y = 360 x +10y + 122 = 3320 x + y = 52 = 0 x +y + z = 33 20 3x + 10y + 12 2 = 340 xty - 520 Hickets of $8 of 10 0 of 12

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