Question 823594
{{{x}}}= number of cases of product 1.
{{{y}}}= number of cases of product 2.
{{{1.65x}}}= cost in $ for {{{x}}} units of product 1.
{{{2.46y}}}= cost in $ for {{{y}}} units of product 2.
{{{1.65x+2.46y}}}= total cost for {{{x}}} units of product 1, and {{{y}}} units of product 2.
according to the vendor's invoice, that is $6115.68, so
{{{1.65x+2.46y=6115.68}}} is our equation.

We have two variables and only one equation.
In general that would mean an infinite number of answers.
However, in this case, only integer, non-negative solutions would work.
That limits the number of solutions.
{{{1.65x+2.46y=6115.68}}}
{{{165x+246y=611568}}}
{{{55x+82y=203856}}}

Once we find one solution, we can increase the {{{x}}} found by {{{82}}} while decreasing the {{{y}}} found by {{{55}}}.
That would change both terms by {{{82*55=4510}}} units and in opposite directions, so that they would still add up to {{{203856}}} .
Since {{{203856}}} divided by {{{4510}}} yields a quotient of {{{45}}} with a remainder of {{{906}}} , there are at least {{{45}}} solutions, and maybe {{{46}}} .
 
If {{{x=y}}} , {{{55x+82x=203856}}} transforms into
{{{137x=203856}}} , {{{x=203856/137}}} , and {{{x=1488}}} .
So one of the possible solutions is {{{x=y=1488}}} .
That is a non-negative integer, so it is allowed to be {{{x}}} and {{{y}}}.
That solution means {{{1488}}} units of each product.

It could be that there were 55 less units of product 2,
meaning {{{1488-55=1433}}} units of product 2,
but 82 more units of product 1,
meaning {{{1488+82=1570}}} .
That would decrease the term {{{82y}}} by {{{82*55=4510}}} ,
but would increase term {{{55x}}} by {{{82*55=4510}}} .
The total would still be {{{203856}}} , with {{{x=1570}}} and {{{y=1433}}} .
We could repeat the procedure a few more times, as long as the decreasing {{{y}}} is still a nonn-egative number.
Since {{{1488}}} divided by {{{55}}} has a quotient of {{{45}}} with a remainder of {{{3}}}, we can do that decreasing {{{y}}} by {{{55}}} a total of {{{27}}} times until we get to {{{y=3}}} .
That means that besides ordered pair (x,y)=(1488,1488), There are {{{27}}} other solutions with lesser values of {{{y}}}, going all the way to {{{x=3702}}} , {{{y=3}}} .
On the other hand, we could decrease {{{x}}} by 82, while increasing {{{y}}} by 55.
That would give us another {{{18}}} solutions, with {{{x}}} decreasing all the way to {{{12}}} because {{{1488 divided by {{{82}}} yields a quotient of {{{18}}} and a remainder of 12.
There is a total of {{{1+27+18=46}}} solutions:
{{{12}}} units of product 1 with {{{2478}}} units of product 2,
{{{94=12+82}}} units of product 1 with {{{2423=2478-55}}} units of product 2,
{{{176}}} units of product 1 with {{{2368}}} units of product 2,
{{{258}}} units of product 1 with {{{2313}}} units of product 2,
and so on, all the way to
{{{3702=12+45*82}}} units of product 1 with {{{3=2368-45*55}}} units of product 2.
I guess we could write them as
{{{x[n]=12+(n-1)*82}}} paired with {{{y[n]=2478-(n-1)*55}}}
foe {{{n}}} integer, such that {{{1<=n<=46}}}