Question 703868
{{{n}}}=number of coats purchased
{{{c}}}=cost of one coat, in $
The cost for {{{n}}} coats, in $, is {{{nc=7200}}}
The cost for {{{8}}} coats, in $, is {{{8c}}}
The {{{n}}} coats were sold for a total of ${{{400n}}}
The profit realized was ${{{400n-7200}}} and that is equal to ${{{8c}}}, so {{{400n-7200=8c}}}
The system of equations to solve is
{{{system(nc=7200,400n-7200=8c)}}}
{{{400n-7200=8c}}} --> {{{50n-900=c}}} (dividing both sides of the equal sign by 8)
So {{{highlight(c=50n-900)}}}
Substituting the expression {{{50n-900}}} for {{{c}}} in {{{nc=7200}}}
we get a quadratic equation:
{{{n(50n-900)=7200}}} --> {{{50n^2-900n=7200}}} --> {{{50n^2-900n-7200=0}}}
Dividing both sides of the equal sign by 50 we simplify that to
{{{n^2-18n-144=0}}} which is easy to solve by factoring,
because we see that {{{n^2-18n-144=(n-24)(n+6)}}}
So we can write the equation as {{{(n-24)(n+6)=0}}}
and realize that {{{n=24}}} and {{{n=-6}}} are solutions to the equation.
Since the number of coats could not be negative, we discard {{{n=-6}}} and our solution is {{{highlight(n=24)}}}.
The number of coats purchased was {{{highlight(24)}}}.
We could even calculate the cost paid for each coat:
Substituting that value in {{{c=50n-900}}} we find
{{{c=50*24-900}}} --> {{{c=1200-900}}} --> {{{c=300}}}.
 
NOTE:
If we cannot see that {{{n^2-18n-144=(n-24)(n+6)}}},
factoring is not an option to solve {{{n^2-18n-144=0}}}.
Then, we have to use the quadratic formula that says that the solutions to
{{{ax^2+bx+c=0}}} are given by {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In the case of {{{n^2-18n-144=0}}} we are using {{{n}}} for {{{x}}},
{{{a=1}}}, {{{b=-18}}} and {{{c=-144}}}, so
{{{n = (-(-18) +- sqrt( (-18)^2-4*1*(-144) ))/(2*1) }}}
{{{n = (18 +- sqrt( 324+576 ))/2 }}}
{{{n = (18 +- sqrt(900))/2 }}} --> {{{n = (18 +- 30)/2 }}}
The solutions are
{{{n=(18 + 30)/2=48/2=24}}} and
{{{n=(18 - 30)/2=(-12)/2=-6}}}