Question 1154401
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Tutor @ikleyn has provided a perfectly good algebraic solution; here is a very different approach....<br>
The math department bought n calculators for d dollars each, for a cost of $240:<br>
{{{nd = 240}}}<br>
They sold (n-1) calculators for (d+5) dollars each, for a total of $300:<br>
{{{(n-1)(d+5) = 300}}}
{{{nd-d+5n-5 = 300}}}
{{{240-d+5n-5 = 300}}}
{{{d = 5n-65}}}
{{{n(5n-65) = 240}}}
{{{n(n-13) = 48}}}
{{{n^2-13n-48 = 0}}}
{{{(n-16)(n+3) = 0}}}
{{{n = 16}}} or {{{n = -3}}}<br>
Obviously the negative answer makes no sense.  So the number of calculators the math department bought was 16.<br>
CHECK:
16 calculators for $240 means $15 each.
16-1 = 15 calculators for $15+$5 = $20 each makes $300.<br>