Question 966201
{{{p}}}= the price/cost per child, in R, based on all children paying the same amount.
{{{n}}}= the number of children in the party.
{{{n*p=18}}}
{{{n-2}}}= the number of children who could pay for the trip.
{{{p+0.10}}}= the amount paid by each child who could pay for the trip.
{{{(n-2)(p+0.10)=18}}}
{{{n*p+0.10n-2p-0.20=18}}}
Substituting {{{18=n*p}}} for {{{n*p}}} , we get
{{{18+0.10n-2p-0.20=18}}}
{{{0.10n-2p-0.20=0}}}
{{{0.10n=2p+0.20}}}
{{{10(0.10n)=10(2p+0.20)}}}
{{{n=20p+2}}}
Substituting the expression {{{20p+2}}} for {{{n}}} in {{{n*p=18}}} , we get
{{{(20p+2)*p=18}}}
{{{20p^2+2p=18}}}
{{{(1/2)(20p^2+2p)=(1/2)(18)}}}
{{{10p^2+p=9}}}
{{{10p^2+p-9=0}}}
That is a quadratic equation that can be easily solved by factoring.
As for all quadratic equations, it can also be solved by "completing the square", or by using the quadratic formula.
To solve by factoring, we look at the coefficients: {{{10}}} , {{{1}}} , and {{{-9}}} .
We look for pairs of factors that multiply to yield
{{{10(-9)=-90}}} and add up to {{{1}}}
{{{90}}} can be written as
{{{1-90=90}}} , {{{2*45=90}}} , {{{3*30=90}}} , {{{5*18=90}}} , {{{6*15=90}}} , and {{{9*10=90}}} .
Making one factor positive and another negative,
we find that {{{10}}} and {{{-9}}} are the pair we are looking for, because
{{{10*(-9)=-90}}} and {{{10+(-9)=1}}} .
We re-wrte {{{10p^2+p-9=0}}} as {{{10p^2+10p-9p-9=0}}} ,
and factor by parts to get
{{{10p(p+1)-9(p+1)=0}}} and {{{(10p-9)(p+1)=0}}} .
The factored form of the equation tells us that one solution comes from
{{{10p-9=0}}}<--->{{{10p=9}}}<--->{{{p=9/10}}}<--->{{{highlight(p=0.90)}}} .
Substituting that value in {{{n*p=18}}} we get
{{{n(0.90)=180}}}<--->{{{n=18/0.90}}}<--->{{{highlight(n=20)}}}
The other solution to the equation is {{{p+1=0}}}<--->{{{p=-1}}} ,
which does not make sense as a per child price/cost for the trip,
so it is not a solution to the problem.