Question 171665
Let x=number of nickels
Then x+4=number of quarters
So the number of dimes would be:
37-x-(x+4)=
37-2x-4=
33-2x----number of dimes

Now we are told that:
(lets deal in pennies)
5x+25(x+4)+10(33-2x)=550  get rid of parens (distributive)
5x+25x+100+330-20x=550 collect like terms
10x+430=550  subtract 430 from each side
10x+430-430=550-430=
10x=120  divide each side by 10
x=12-----------------number of nickels
x+4=12+4=16-------------number of quarters
33-2x=33-24=9---------------number of dimes

CK
5*12+25*16+9*10=550
60+400+90=550
550=550
We could also solve this problem using three unknowns:
x=number of nickels
y=number of dimes
z=number of quarters
x+y+z=37-----------------------eq1
5x+10y+25z=550------------------eq2
z=x+4----------------------------eq3
Substitute z=x+4 into eq1 and eq2:
2x+y=33-------------eq1a
30x+10y=450-----------eq2a
Multiply eq1a by 15 and subtract eq2a from it:
5y=45
y=9---------------number of dimes


Hope this helps---ptaylor