algebra word problem- help!!?
A local comedy club put on a play. Adult tickets cost $6.50 each while children's tickets sold for $3.75. There were 290 people who attended the performances. The total receipts from the sale of tickets amounted to $1656.75 how many adults and children attended?
Comments
6.50A + 3.75C = 1656
A + C = 290
A = 290 -C
6.50(290 - C) + 3.75C = 1656.75
1885 - 6.50C + 3.75C = 1656.75
-2.75C = -228.25
C = 83
A = 290 - 83 = 207
Adults = a , and children = c.
a + c = 290, so a = 290 -c
6.50a + 3.75 c = 1656.75
6.50(290 - c ) +3.75 c = 1656.75 ( substitute a in the second equation by 290-c from the first equation).
1885 - 6.5c + 3.75c =1656.75
-2.75 c =1656.75 - 1885
c =83 If 83 children, then 207 adults ( 83*3.75 ) + (207*6.50)=1656.75