Question 1029850
We start by giving variable names that will work as placeholders for the values we need to find:
{{{p}}}= number of plain burgers,
{{{c}}}= number of cheeseburgers,
{{{d}}}= number of double cheeseburgers.
(Some people like to use {{{x}}} , {{{y}}}, and {{{z}}} 
but I like variable names that remind of their meaning).
 
All the burgers require {{{1}}} bun per burger,
so {{{p+c+d}}} buns are needed,
and {{{14*8}}} buns are available(14 packs of 8 buns per pack).
That gives us the equation
{{{p+c+d=14*8}}} , or {{{p+c+d=112}}} .
 
The plain burgers, and the cheeseburgers require one patty each,
but each double cheeseburger required 2 patties,
so the number of frozen hamburger patties required is
{{{p+c+2d}}} ,
and {{{12*12}}} patties area available (12 packs, with 12 frozen hamburger patties per pack).
That gives us the equation
{{{p+c+2d=12*12}}} , or {{{p+c+2d=144}}} .
 
A plain hamburger requires no cheese,
while a cheeseburger requires one slice of cheese,
and a double cheeseburger required 2 slices,
so {{{c+2d}}} slices of cheese will be needed.
There area {{{3*24}}} slices of cheese available (3 packs, with 24 slices per pack),
and that gives us the equation
{{{c+2d=3*24}}} or {{{c+2d=72}}} .
 
The three equation form the easy system of linear equations
{{{system(p+c+d=112,p+c+2d=144,c+2d=72)}}} .
Subtracting the first equation from the second, we get
{{{highlight(d=32)}}} .
Substituting that value in the third equation we get
{{{c+2*32=72}}} --> {{{c+64=72}}} --> {{{c=72-64}}} --> {{{highlight(c=8)}}} .
Substituting the values found for {{{c}}} and {{{d}}} into the first equation we get
{{{p+8+32=112}}} --> {{{p+40=112}}} --> {{{p=112-40}}} --> {{{highlight(p=72)}}} .
 
plain hamburgers  72 
cheeseburgers  8 
double cheeseburgers  32.