Question 170726
Let c=cost per individual (the cost before the students failed to arrive) and n=number of students



So the total cost is divided among n people to get


{{{c=500/n}}}



Since 5 failed to attend, this means that {{{n-5}}} students showed up. Now the new cost is


{{{c+5=500/(n-5)}}}



{{{c+5=500/(n-5)}}} Start with the given equation.



{{{500/n+5=500/(n-5)}}} Plug in {{{c=500/n}}}



{{{500(n-5)+5n(n-5)=500n}}} Multiply every term by the LCD {{{n(n-5)}}}. This cancels out every denominator.



{{{500n-2500+5n^2-25n=500n}}} Distribute



{{{500n-2500+5n^2-25n-500n=0}}} Subtract {{{500n}}} from both sides.



{{{5n^2-25n-2500=0}}} Subtract {{{500n}}} from both sides.



Let's use the quadratic formula to solve for n



{{{n = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{n = (-(-25) +- sqrt( (-25)^2-4(5)(-2500) ))/(2(5))}}} Plug in  {{{a=5}}}, {{{b=-25}}}, and {{{c=-2500}}}



{{{n = (25 +- sqrt( (-25)^2-4(5)(-2500) ))/(2(5))}}} Negate {{{-25}}} to get {{{25}}}. 



{{{n = (25 +- sqrt( 625-4(5)(-2500) ))/(2(5))}}} Square {{{-25}}} to get {{{625}}}. 



{{{n = (25 +- sqrt( 625--50000 ))/(2(5))}}} Multiply {{{4(5)(-2500)}}} to get {{{-50000}}}



{{{n = (25 +- sqrt( 625+50000 ))/(2(5))}}} Rewrite {{{sqrt(625--50000)}}} as {{{sqrt(625+50000)}}}



{{{n = (25 +- sqrt( 50625 ))/(2(5))}}} Add {{{625}}} to {{{50000}}} to get {{{50625}}}



{{{n = (25 +- sqrt( 50625 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{n = (25 +- 225)/(10)}}} Take the square root of {{{50625}}} to get {{{225}}}. 



{{{n = (25 + 225)/(10)}}} or {{{n = (25 - 225)/(10)}}} Break up the expression. 



{{{n = (250)/(10)}}} or {{{n =  (-200)/(10)}}} Combine like terms. 



{{{n = 25}}} or {{{n = -20}}} Simplify. 



Since a negative amount of people doesn't make any sense, this means that the only answer is {{{n = 25}}}



So 25 people were going to attend but only 20 actually attended.