Question 1011499
All approaches have their merits, and Edwin's strategy works well in all cases.
However, in this particular case,
combining that strategy with a bit of trial and error may lead to the answer faster, or easier, or in a more intuitive way.
{{{d}}}= number of ducks
{{{p}}}= number of pigs
{{{c}}}= number of chicks
{{{d+p+c=100}}}= total number of animals
{{{0.5c+3p+10d=100}}}= total amount spend, in $.
 
{{{d+p+c=100}}}<-->{{{d=100-p-c}}}
{{{system(d=100-p-c,0.5c+3p+10d=100)}}}-->{{{system(d=100-p-c,0.5c+3p+10(100-p-c)=100)}}}-->{{{system(d=100-p-c,0.5c+3p+1000-10p-10c=100)}}}-->{{{system(d=100-p-c,900=7p+9.5c)}}}-->{{{system(d=100-p-c,14p+19c=1800)}}}-->{{{system(d=100-p-c,p=(1800-19c)/14)}}} .
Choosing intelligently, we can assign values to {{{c}}} between {{{0}}} and {{{100}}} ,
and find integers {{{p}}} and {{{d}}} between {{{0}}} and {{{100}}} .
Since {{{1800/14=128&8/14}}} , {{{c=0}}} does yield an integer value for {{{p}}} ,
but {{{c=10}}} does, yielding {{{p=115}}} ,
and every {{{14}}} unit increase in {{{c}}} results in a {{{19}}} unit reduction in {{{p}}} ,
so {{{c=24}}} yields {{{p=96}}} , and
{{{c=38}}} yields {{{p=77}}} .
So, we could try {{{c=24}}} , {{{c=38}}} , {{{c=52}}} ,
and a few more, until we get to a {{{c>100}}} .
How far can we continue?
Since {{{1800/19=94&14/19}}} ,
{{{system(c=94,p=1)}}} is as far as we can go.
{{{system(c=94,p=1,d=100-p-c)}}}--->{{{system(c=94,p=1,d=100-94-1)}}}--->{{{highlight(system(c=94,p=1,d=5))}}} is one solution.
Before getting to {{{c=94}}} we hit {{{c=80}}} , and {{{c=66}}} .
{{{system(c=80,p=(1800-19c)/14,d=100-p-c)}}}--->{{{system(c=80,p=(1800-19*80)/14,d=100-p-c)}}}--->{{{system(c=80,p=(1800-1520)/14,d=100-p-c)}}}--->{{{system(c=80,p=280/14,d=100-p-c)}}}--->{{{system(c=80,p=20,d=100-80-20)}}}--->{{{highlight(system(c=80,p=20,d=0))}}} is another solution.
{{{c=66}}} and lesser values do not yield solutions because they cause {{{d}}} to be negative.