Question 1071362
Translation into Algebreze:
{{{x+y=8}}} because "there are eight notes altogether."
{{{5x+10y=55}}} because "their total value is N55."
 
SOLUTION BY SUBSTITUTION:
{{{system(x+y=8,5x+10y=55)}}} --> {{{system(x=8-y,5x+10y=55)}}} --> {{{system(x=8-y,5(8-y)+10y=55)}}} --> {{{system(x=8-y,40-5y+10y=55)}}} --> {{{system(x=8-y,40+5y=55)}}} --> {{{system(x=8-y,5y=55-40)}}} --> {{{system(x=8-y,5y=15)}}} --> {{{system(x=8-y,y=15/5)}}} --> {{{system(x=8-y,y=3)}}} --> {{{system(x=8-3,y=3)}}} --> {{{highlight(system(x=5,y=3))}}}
 
A SMARTER WAY:
Simplify first, like this
{{{5x+10y=55}}} --> {{{(5x+10y)/5=55/5}}} --> {{{x+2y=11}}}
 
THE FIFTH GRADER'S SOLUTION:
N5 notes are better because if you only carry one N5 note,
I cannot be tempted to spend N10 at one time.
I should change those N10 notes,
trading each N10 note for 2  N5 notes.
I will have the same amount of money,
but for each N10 note I get changed, I will have one more note.
I will still have N55,
but I will have {{{55/5=11}}} notes rather than {{{8}}} .
That would mean that I would have changed {{{11-8=3}}} N10 notes.
So I have {{{highlight(3)}}} N10 notes,
and {{{8-3=highlight(5)}}} N5 notes.