SOLUTION: your company reached its goal so you decide to buy dinner for your 45 employees. the chicken dinners are $9, salmon dinner is $12, and steak dinner $15. the number of chicken dinne

Algebra ->  Rational-functions -> SOLUTION: your company reached its goal so you decide to buy dinner for your 45 employees. the chicken dinners are $9, salmon dinner is $12, and steak dinner $15. the number of chicken dinne      Log On


   



Question 1039757: your company reached its goal so you decide to buy dinner for your 45 employees. the chicken dinners are $9, salmon dinner is $12, and steak dinner $15. the number of chicken dinners you order is half the sum of the number of salmon and steak dinners. the total cost of all meals is $525, how many of each type of dinner do you buy?
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please help!! thank you!!!

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = number of chicken dinners bought
Let +b+ = number of salmon dinners bought
Let +c+ = number of steak dinners bought
-----------------------------------------
(1) +a+=+%281%2F2%29%2A%28+b+%2B+c+%29+
(2) +a+%2B+b+%2B+c+=+45+
(3) +9a+%2B+12b+%2B+15c+=+525+
--------------------------
There are 3 equations and 3 unknowns, so it's solvable
(1) +2a+=+b+%2B+c+
Substitute (1) into (2)
(2) +a+%2B+2a++=+45+
(2) +3a+=+45+
(2) +a+=+15+
-------------------
(2) +15+%2B+b+%2B+c+=+45+
(2) +b+%2B+c+=+30+
and
(3) +9%2A15+%2B+12b+%2B+15c+=+525+
(3) +135+%2B+12b+%2B+15c+=+525+
(3) +12b+%2B+15c+=+390+
(3) +4b+%2B+5c+=+130+
------------------------
Multiply both sides of (2) by +4+
and subtract (2) from (3)
(3) +4b+%2B+5c+=+130+
(2) +-4b+-+4c+=+-120+
----------------------
+c+=+10+
and
(2) +b+%2B+c+=+30+
(2) +b+%2B+10+=+30+
(2) +b+=+20+
--------------------
You bought:
15 chicken dinners
20 salmon dinners
10 steak dinners
--------------------
check:
(3) +9a+%2B+12b+%2B+15c+=+525+
(3) +9%2A15+%2B+12%2A20+%2B+15%2A10+=+525+
(3) +135+%2B+240+%2B+150+=+525+
(3) +525+=+525+
OK