SOLUTION: At a thearer, 400 tickets were sold. The ticket prices were $8, $10, and $12. The total income from ticket sales was $3700. How many tickets of each type were sold if the combined

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: At a thearer, 400 tickets were sold. The ticket prices were $8, $10, and $12. The total income from ticket sales was $3700. How many tickets of each type were sold if the combined       Log On


   



Question 574467: At a thearer, 400 tickets were sold. The ticket prices were $8, $10, and $12. The total income from ticket sales was $3700. How many tickets of each type were sold if the combined number of $8 and $10 tickets sold was seven times the number of $12 tickets sold.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = number of $8 tickets
Let b = number of $10 tickets
Let +c = number of $12 tickets
given:
(1) +a+%2B+b+%2B+c+=+400+
(2) +8a+%2B+10b+%2B+12c+=+3700+
(3) +a+%2B+b+=+7c+
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This is 3 equations and 3 unknowns, so
it should be solvable
Substitute (3) into (1)
(1) +7c+%2B+c+=+400+
(1) +8c+=+400+
(1) +c+=+50+
Plug this value into (2)
(2) +8a+%2B+10b+%2B+12%2A50+=+3700+
(2) +8a+%2B+10b+%2B+600+=+3700+
(2) +8a+%2B+10b+=+3100+
(2) +4a+%2B+5b+=+1550+
and
(1) +a+%2B+b+%2B+c+=+400+
(1) +a+%2B+b+%2B+50+=+400+
(1) +a+%2B+b+=+350+
Multiply both sides of (1) by +4+
(1) +4a+%2B+4b+=+1400+
Subtract (1) from (2)
(2) +4a+%2B+5b+=+1550+
(1) +-4a+-+4b+=+-1400+
+b+=+150+
and, since
(3) +a+%2B+b+=+7c+
(3) +a+%2B+150+=+7%2A50+
(3) +a+=+350+-+150+
(3) +a+=+200+
There are 200 $8 tickets
There are 150 $10 tickets
There are 50 $12 tickets
check:
(2) +8%2A200+%2B+10%2A150+%2B+12%2A50+=+3700+
(2) +1600+%2B+1500+%2B+600+=+3700+
(2) +3700+=+3700+
OK