Question 1013111
{{{x}}}= number of cups sold
{{{y}}}= number of cones sold
{{{0.6x}}}= money (in $) made by selling {{{x}}} cups
{{{2y}}}= money (in $) made by selling {{{y}}} cones
{{{0.6x+2y=200}}}= total money (in $) made by selling {{{x}}} cups and {{{y}}} cones.
I would expect another piece of information, such as the fact that they served a certain number of customers.
 
IF the problem stated that {{{135}}} servings (between cones and cups) were sold
we would write that {{{x+y=135}}} , and we would find {{{x}}} and {{{y}}} .
by solving the system of linear equations {{{system(x+y=135,0.6x+2y=200)}}}
{{{system(x+y=135,0.6x+2y=200)}}}--->{{{system(y=135-x,0.6x+2y=200)}}}--->{{{system(y=135-x,0.6x+2(135-x)=200)}}}--->{{{system(y=135-x,0.6x+270-2x=200)}}}--->{{{system(y=135-x,0.6x-2x=200-270)}}}
--->{{{system(y=135-x,-1.4x=-70)}}}--->{{{system(y=135-x,x=(-70)/(-1.4))}}}--->{{{system(y=135-x,x=50)}}}--->{{{system(y=135-50,x=50)}}}--->{{{system(y=85,x=50)}}}
 
Since {{{0.6x+2y=200}}} is the only equation you can write from the information given,
we should expect to find more than one solution.
Only fact that {{{x}}} and {{{y}}} must both be integers keeps the problem from having {{{infinity}}} solutions.
{{{0.6x+2y=200}}}<-->{{{0.3x+y=100}}}<-->{{{y=100-0.3x}}}
For {{{y}}} to be an integer, we need {{{0.3x}}}} to be an integer,
and for {{{0.3x=3x/10}}} to be an integer, {{{x}}} must be a multiple of {{{10}}} .
So we take non-negative values for {{{x}}} that can be divided by {{{10}}} ,
calculate {{{y=100-0.3x}}} for each {{{x}}} value,
and tabulate all the many solutions we find:
{{{matrix(2,12,
x,0,1,2,3,4,5,".....",300,310,320,330,
y,100,97,94,91,88,85,".....",10,7,4,1)}}}