Algebra is probably the only way.

3 camera and 5 printer is 4280
5 camera and 4 printer is 4620

We try and make the number of printers the same in both cases (which is essentially algebra).

A common multiple of 5 and 4 is 20. We try and multiply the printers to 20.

If 3 cameras and 5 printers is 4280, then 4 sets of this would give 12 cameras and 20 printers 17120.

If 5 cameras and 4 printers is 4620, then 5 sets of this would give 25 cameras and 20 printers 23100.

So now,

12 cameras + 20 printers = 17120
25 cameras + 20 printers = 23100

The difference between the two cases is 13 cameras and the price difference is 5980.

Each camera must have cost 460.

Remember that 3 cameras and 5 printers cost 4280.

3 cameras is 1380. So 5 printers is 2900 and 1 printer is 580.

This technique is known as the elimination method and will be seen again in Sec 2 or Sec 3.