Problem solving need help
- Thread starter Brunswick
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HallsofIvy
Elite Member
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Let x be the cost of a printer and y be the cost of a scanner.
"3 printers in 1 scanner cost $842.50 dollars."
So 3x+ y= 842.50
"1 printer and 2 scanner cost $690."
So x+ 2y= 690.00.
Multiply the second equation by 3: 3x+ 6y= 2070. Subtract the first equation 3x+ y= 842.50 from that. The "3x" terms in each equation cancel leaving
6y- y= 5y= 2070- 842.50= 1227.50. Solve 5y= 1227.50 to find the cost of one scanner.
"3 printers in 1 scanner cost $842.50 dollars."
So 3x+ y= 842.50
"1 printer and 2 scanner cost $690."
So x+ 2y= 690.00.
Multiply the second equation by 3: 3x+ 6y= 2070. Subtract the first equation 3x+ y= 842.50 from that. The "3x" terms in each equation cancel leaving
6y- y= 5y= 2070- 842.50= 1227.50. Solve 5y= 1227.50 to find the cost of one scanner.
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- Feb 4, 2004
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- 16,582
Since you posted this to "Arithmetic", I'll assume you don't yet know about variables.
If you get two of the first order, how many printers and scanners will you have? How much will this double order cost?
If you subtract the second order from the doubled first order, how many printers and scanner will you have? How much will the result cost?
Where does this lead?
