quote:

Originally posted by Acid Junkie that is too easy a solution u've got there. give the man something he can show off with dude, if u want to get an A++ in ur class, show them sometingh like this. this is how to solve the problem using 3(!) variables. let x be the number of red pens you can buy and y the number of blue pens. then, given that u can only spend $4.95=495 cents, we have 17x+13y≤495 x≥0, y≥0 since the difference of the number of both kinds of pens cannot exceed 5, we have |x-y|≤5 we need to find the biggest value that x+y=z can take so what we have here is a nice mixed system with 3 variables 17x + 13y≤495 |x-y|≤5 x+y=z 0≤x≤z 0≤y≤z applying a few substitutions for y (y=z-x), we get 4x+13z≤495 |2x-z|≤5 which is the same as x≤(495-13z)/4 (z-5)/2≤x≤(z+5)/2 considering that 0≤x≤z, we can rewrite the above system as z≥0 (z+5)/2≥0 0≤(495-13z)/4 <=> 495-13z≥0 <=> z≤495/13 (z-5)/2≤(495-13z)/4 <=> 2(z-5)≤(495-13z) <=> 15z≤505 <=> z≤505/15 or as 0≤z -5≤z z≤495/13 z≤505/15 or as 0≤z≤505/15=101/3=33+2/3 which means the greatest integer z can be is 33 so we need to find the smallest x possible when z=33. using 33 for z, we get another system with only one variable x≤16.5 14≤x≤19 x≤33 the smallest solution of which is 14. same answer

having a third variable is redundant.