Question 834512
{{{a}}}= number of days factory A needs to work to exactly fill that order.
{{{b}}}= number of days factory B needs to work to exactly fill that order.
 
During {{{a}}} days, factory A can make {{{25a}}} standard widgets.
During {{{b}}} days, factory B can make {{{150b}}} standard widgets.
To make the 500 standard widgets {{{a}}} and {{{b}}} have to satisfy the equation
{{{25a+150b=500}}} .
We can simplify that equation a bit by dividing both sides of the equal sign by 25.
{{{(25a+150b)/25=500/25}}}
{{{25a/25+150b/25=500/25}}}
{{{highlight(a+6b=20)}}}
 
During {{{a}}} days, factory A can make {{{50a}}} deluxe widgets.
During {{{b}}} days, factory B can make {{{75b}}} deluxe widgets.
To make the 325 deluxe widgets {{{a}}} and {{{b}}} have to satisfy the equation
{{{50a+75b=325}}} .
We can simplify that equation a bit by dividing both sides of the equal sign by 25. 
{{{50a+75b=500}}}
{{{50a/25+75b/25=325/25}}}
{{{highlight(2a+3b=13)}}}
 
We can set up the simpler system
{{{highlight(system(a+6b=20,2a+3b=13))}}}
or we could have stopped before simplifying the equations and end up with
{{{highlight(system(a+6b=20,50a+75b=500))}}}
 
IF WE HAD TO SOLVE THE SYSTEM:
There are several ways to solve the system.
We could solve {{{a+6b=20}}} for {{{a}}} to get an expression that is equal to {{{a}}} ,
{{{a+6b=20}}} ---> {{{a=20-6b}}}
and the substitute the expression {{{20-6b}}} for {{{a}}} in the other equation:
{{{system(a=20-6b,2a+3b=13)}}} --> {{{2(20-6b)+3b=13}}}
Then we solve the resulting equation for {{{b}}}
{{{2(20-6b)+3b=13}}}
{{{2*20-2*6b+3b=13}}}
{{{40-12b+3b=13}}}
{{{40-9b=13}}}
{{{-9b=13-40}}}
{{{-9b=-27}}}
{{{b=(-27)/(-9)}}}
{{{highlight(b=3)}}}
Then we would substitute the value found for {{{b}}} into {{{a=20-6b}}} and get
{{{a=20-6*3}}}
{{{a=20-18}}}
{{{highlight(a=2)}}}