SOLUTION: lAST YEAR THE COMPUTER LAB PURCHASED 2 DIFFERENT TYPES OF COMPUTERS, 8 TYPE A PC'S AND 12 TYPE B PC'S. THE COST OF THESE COMPUTERS WAS $21,200. THIS YEAR THE COMPUTER COMPANY SAID

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: lAST YEAR THE COMPUTER LAB PURCHASED 2 DIFFERENT TYPES OF COMPUTERS, 8 TYPE A PC'S AND 12 TYPE B PC'S. THE COST OF THESE COMPUTERS WAS $21,200. THIS YEAR THE COMPUTER COMPANY SAID      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 697881: lAST YEAR THE COMPUTER LAB PURCHASED 2 DIFFERENT TYPES OF COMPUTERS, 8 TYPE A PC'S AND 12 TYPE B PC'S. THE COST OF THESE COMPUTERS WAS $21,200. THIS YEAR THE COMPUTER COMPANY SAID IT WOULD SELL COMPUTERS TO THE LAB AT THE SAME COST PER COMPUTER. THIS YEAR THE LAB PURCHASED 12 TYPE A PC'S AND 5 TYPE B PC'S FOR $16,200. HOW MUCH DID EACH TYPE A PC COST? HOW MUCH DID EACH TYPE B PC COST?
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
Let A = Cost of type A PC
Let B = Cost of type B PC
Equation 1: 8A+%2B+12B+=+21200
Equation 2: 12A+%2B+5B+=+16200
I am going to solve this by using the substitution method.
Note that a common mutiple of 8A and 12A is 24A.
Start by setting both equations to zero
Equation 1: 8A+%2B+12B+-+21200+=+0
Equation 2: 12A+%2B+5B+-+16200+=+0
Multiply equation 1 by 3 to get 24A
Equation 1: 24A+%2B+36B+-+63600+=+0
Multiply equation 2 by 2 to get 24A
Equation 2: 24A+%2B+10B+-+32400+=+0
Now set both equations equal to each other
24A+%2B+36B+-+63600+=+24A+%2B+10B+-+32400
Subtract 24A from both sides
36B+-+63600+=+10B+-+32400
Add 63600 to both sides
36B+=+10B+%2B+31200
Subtract 10B from both sides
26B+=+31200
Divide both sides by 26
highlight%28B+=+1200%29
-----------------------
Now plug 1200 into one of the given equations for B
Equation 1: 8A+%2B+12B+=+21200
8A+%2B+12%2A%281200%29+=+21200
8A+%2B+14400+=+21200
Subtract 14400 from btoh sides
8A+=+6800
Divide both sides by 8
highlight_green%28A+=+850%29