SOLUTION: Theresa sold magazine subscriptions with three prices: $12, $13, and $22. She sold 1 fewer of the $12 subscriptions than of the $13 subscriptions and sold a total of 40 subscripti

Algebra ->  Human-and-algebraic-language -> SOLUTION: Theresa sold magazine subscriptions with three prices: $12, $13, and $22. She sold 1 fewer of the $12 subscriptions than of the $13 subscriptions and sold a total of 40 subscripti      Log On


   



Question 582513: Theresa sold magazine subscriptions with three prices: $12, $13, and $22. She sold 1 fewer of the $12
subscriptions than of the $13 subscriptions and sold a total of 40 subscriptions. If her total sales
amounted to $643, how many $22 subscriptions did Theresa sell?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = number of $12 ones she sold
Let +b+ = number of $13 ones she sold
Let +c+ = number of $22 ones she sold
given:
(1) +a+=+b+-+1+
(2) +a+%2B+b+%2B+c+=+40+
(3) +12a+%2B+13b+%2B+22c+=+643+
This is 3 equations and 3 unknowns, so it's solvable
---------
Multiply both sides of (2) by 22 and
subtract (3) from (2)
(2) +22a+%2B+22b+%2B+22c+=+880+
(3) +-12a+-+13b+-+22c+=+-643+
+10a+%2B+9b+=+237+
Substitute (1) into this result
+10%2A%28+b+-+1+%29+%2B+9b+=+237+
+10b+-+10+%2B+9b+=+237+
+19b+=+247+
+b+=+13+
and, since
(1) +a+=+b+-+1+
(1) +a+=+13+-+1+
(1) +a+=+12+
and
(2) +a+%2B+b+%2B+c+=+40+
(2) +12+%2B+13+%2B+c+=+40+
(2) +25+%2B+c+=+40+
(2) +c+=+40+-+25+
(2) +c+=+15+
She sold 15 of the $22 subscriptions
check answer:
(3) +12a+%2B+13b+%2B+22c+=+643+
(3) +12%2A12+%2B+13%2A13+%2B+22%2A15+=+643+
(3) +144+%2B+169++%2B+330+=+643+
(3) +643+=+643+
OK