SOLUTION: A company orders two types of parts , brass and steel . One shipment contains 3 brass and 10 steel parts and costs $48. A second shipment contains 7 brass and 4 steel parts and co

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A company orders two types of parts , brass and steel . One shipment contains 3 brass and 10 steel parts and costs $48. A second shipment contains 7 brass and 4 steel parts and co      Log On


   

Question 629810: A company orders two types of parts , brass and steel . One shipment contains 3 brass and 10 steel parts and costs $48. A second shipment contains 7 brass and 4 steel parts and costs $54. Find the cost of each type of part .
Found 3 solutions by solver91311, richwmiller, stanbon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Let represent the cost of brass and represent the cost of steel.





Solve the 2X2 system for

John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
3b+10s=48;
7b+4s=54

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A company orders two types of parts , brass and steel . One shipment contains 3 brass and 10 steel parts and costs $48. A second shipment contains 7 brass and 4 steel parts and costs $54. Find the cost of each type of part .
------
Equations:
3b + 10s = 48 dollars
7b + 4s = 54 dollars
-----------------------
Multiply thru 1st by 7; thru 2nd by 3
----
21b + 70s = 7*48
21b + 12s = 3*54
---------
Subtract and solve for "s":
58s = 174
s = $3 (price of one steel part)
----
Solve for "b":
3b + 10*s = 48
3b + 30 = 48
b = $6 (price of one brass part)
===================================
Cheers,
Stan H.
==============================