Question 947387
I guess your "loonies" are Canadian $1 coins. (I do not recall ever seeing one, but I have only visited Canada in very short trips).
 
IF YOU HAVE STUDIED SYSTEMS OF LINEAR EQUATIONS:
{{{x}}}= number of loonies
{{{y}}}= number of quarters
"I have 200 loonies and quarters" translates as
{{{x+y=200}}} .
Since
{{{x}}}= value of {{{x}}} loonies in Canadian dollars, and
{{{0.25y=y/4}}}= value of {{{y}}} quarters in Canadian dollars,
if the total of both types of coins amounts to 91.25,
{{{x+0.25y=91.25}}} or {{{x+y/4=91.25}}} .
We can turn that equation into a prettier equivalent equation
by multiplying both sides of the equal sin times {{{t}}}o get
{{{4(x+0.25y)=4*91.25}}}<--->{{{4x+y=365}}}
 
Now we have
{{{system(x+y=200,4x+y=365)}}}-->{{{system(y=200-x,4x+y=365)}}}-->{{{system(y=200-x,4x+200-x=365)}}}-->{{{system(y=200-x,3x+200=365)}}}-->{{{system(y=200-x,3x=365-200)}}}-->{{{system(y=200-x,3x=165)}}}-->{{{system(y=200-x,x=165/3)}}}-->{{{system(y=200-x,x=55)}}}-->{{{system(y=200-55,x=55)}}}-->{{{highlight(system(y=145,x=55))}}}
 
IF YOU HAVE NOT STUDIED SYSTEMS OF LINEAR EQUATIONS:
{{{x}}}= number of loonies
"I have 200 loonies and quarters" translates as
{{{200-x}}}= number of quarters.
Since
{{{x}}}= value of {{{x}}} loonies in Canadian dollars, and
{{{0.25(200-x)}}}= value of {{{200-x}}} quarters in Canadian dollars,
if the total of both types of coins amounts to 91.25,
{{{x+0.25(200-x)=91.25}}} .
Then,
{{{x+0.25(200-x)=91.25}}}-->{{{x+50-0.25x=91.25}}}-->{{{x-0.25x+50=91.25}}}-->{{{x(1-0.25)+50=91.25}}}-->{{{0.75x+50=91.25}}}-->{{{0.75x=91.25-50}}}-->{{{0.75x=41.25}}}-->{{{x=41.25/0.75}}}-->{{{highlight(x=55)}}},
and {{{200-x=200-55}}}-->{{{200-x=highlight(145)}}} .
 
Either way, you have {{{highlight(55)}}} loonies and {{{highlight(145)}}} quarters.