Arthur D. answered 11/15/14
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A=# of $25 tickets
B=# of $35 tickets
C=# of $50 tickets
A+B+C=40 but he wants to buy twice as many $35(B) tickets as $25(A) tickets
A+2A+C=40
3A+C=40
$25A+$35(2A)+$50C=$1670
$25A+$70A+$50C=$1670
$95A+$50C=$1670
3A+C=40 so C=40-3A
substitute
$95A+$50(40-3A)=$1670
95A+2000-150A=1670
-55A+2000=1670
-55A=1670-2000
-55A=-330
55A=330
A=330/55
A=6 $25 tickets
B=12 $35 tickets
40-6-12=40-18=22
C=22 $50 tickets
check: 6*$25+12*$35+22*$50=$150+$420+$1100=$1670