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
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-> 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
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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) (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:
Equation 2:
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:
Equation 2:
Multiply equation 1 by 3 to get 24A
Equation 1:
Multiply equation 2 by 2 to get 24A
Equation 2:
Now set both equations equal to each other
Subtract 24A from both sides
Add 63600 to both sides
Subtract 10B from both sides
Divide both sides by 26
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Now plug 1200 into one of the given equations for B
Equation 1:
Subtract 14400 from btoh sides
Divide both sides by 8