Answer:
There are 200 $8 tickets
There are 150 $10 tickets
There are 50 $12 tickets
Step-by-step explanation:
Let a = number of $8 tickets
Let b = number of $10 tickets
Let +c = number of $12 tickets
given:
(1) a+b+c = 400
(1) 8a+10b+12c = 3700
(1) a+b = 7c
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This is 3 equations and 3 unknowns, so
it should be solvable
Substitute (3) into (1)
(1) 7c+c = 400
(1) 8c= 400
(1) c = 50
Plug this value into (2)
(2) 8a+10b+12*50= 3700
(2) 8a+10b+600= 3700
(2) 8a+10b= 3100
(2) 4a+5b= 1550
and
(1) a+b+c= 400
(1) a+b+50= 400
(1) a+b= 350
Multiply both sides of (1) by +4+
(1) 4a+4b= 1400
Subtract (1) from (2)
(2) 4a+5b+=+1550+
(1) -4a-4b= -1400
b= 150
and, since
(3) a+b= 7c
(3) a+150= 7*50
(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*200+10*150+12*50= 3700
(2) 1600+1500+600= 3700
(2) 3700= 3700