Dorris is trying to decide between two laser printers. Both have similar features and warranties so price is the determining factor. Printer A costs $159 and printing costs are approximately $004 per page Printer B costs $19 : and printing costs are approximately $0.02 per page. How many pages need to be printed for the cocit afi the turc printers to be the same? pages need to be printed for the cost of the two printers to be equal.

Question

Dorris is trying to decide between two laser printers. Both have similar features and warranties so price is the determining factor. Printer A costs  $159 and printing costs are approximately  $004 per page Printer B costs  $19 : and printing costs are approximately  $0.02 per page. How many pages need to be printed for the cocit afi the turc printers to be the same? pages need to be printed for the cost of the two printers to be equal.

Bytelearn · Accepted Answer

Define total cost for Printer A: Let's denote the number of pages printed as 'p'. The total cost of using Printer A for 'p' pages is the initial cost of the printer plus the cost per page times the number of pages. For Printer A, this is $159 + $0.04p.Define total cost for Printer B: Similarly, the total cost of using Printer B for 'p' pages is the initial cost of the printer plus the cost per page times the number of pages. For Printer B, this is $19 + $0.02p.Set up equation for equal costs: We want to find the point where the costs are equal, so we set the total costs for both printers equal to each other: $159 + $0.04p = $19 + $0.02pIsolate terms with 'p': To solve for 'p', we first subtract $19 from both sides of the equation to isolate the terms with 'p' on one side: $159 - $19 + $0.04p = $19 - $19 + $0.02p $140 + $0.04p = $0.02pCombine 'p' terms: Next, we subtract $0.02p from both sides to get all the 'p' terms on one side: $140 + $0.04p - $0.02p = $0.02p - $0.02p $140 + $0.02p = 0Solve for 'p': Now, we divide both sides by $0.02 to solve for 'p': $140 / $0.02 = $0.02p / $0.02 7000 = p

Full solution

More problems from Weighted averages: word problems

Related topics