Question 255012
bill = $239.25


let x = number of students.


let y = amount each student paid.


if all students paid the bill, then:


{{{239.25 / x = y}}} (first equation)


if all but 4 paid the bill, then:


{{{239.25 / (x-4) = y+1}}} (second equation)


substitute for y in second equation by the value for y in first equation to get:


{{{239.25 / (x-4) = (239.25/x) + 1}}}


you had one equation in 2 unknowns.   the substitution allows you to have 1 equation in 1 unknown.


multiply both sides of this equation by (x-4) to get:


{{{239.25 = (x-4) * ((239.25/x) + 1)}}}


multiply the factors to get:


{{{239.25 = 239.25 + x - 4*(239.25/x) - 4}}}


subtract 239.25 from both sides of the equation to get:


{{{0 = x - 4*(239.25/x) - 4}}}


remove parentheses to get:


{{{0 = x - (957)/x - 4}}}


multiply both sides of equation by x to get:


{{{0 = x^2 - 957 - 4*x}}}


rearrange terms to get:


{{{x^2 - 4*x - 957 = 0}}}


use quadratic formula to solve for x to get:


x = {{{(-(-4) +- sqrt(16^2 - (4*1*(-957))))/(2)}}}


this becomes:


x = {{{(4 +- sqrt(16^2 + 3828))/(2)}}} which becomes:


x = {{{(4 +- sqrt(3844))/2}}} which becomes:


x = {{{(4 +- 62)/2}}} which becomes:


{{{x = (4+62)/2 = 66/2 = 33 }}}


or:


{{{x = (4-62)/2 = -58/2 = -29}}}


since x has to be positive, then the solution narrows down to:


{{{x = 33}}}.


when 33 students pay the bill, the bill is $239.25 / 33 = $7.25 apiece.


when 29 students pay the bill (33-4 = 29), the bill is $239.25 / 29 = $8.25.