Question 163739
Let x=number of pennies
y=number of nickels
And z=number of dimes
(we'll deal in pennies)
x+5y+10z=165----------------------------eq1
y=x+z-----------------------------------------eq2
subtract eq2 from eq1 and we get:

6y+9z=165  subtract 9z from each side
6y=165-9z  divide each side by 6
y=(165-9z)/6=(55-3z)/2-------------eq3  
eq3 shows the relationship between number of nickels and number of dimes.  Now, we know the following about this problem:

(1)----We cannot have fractions of coins;  we must deal in whole numbers
(2)----And we cannot have negative coins;  we must deal in positive numbers

By inspection, we see immediately that z, the number of dimes, cannot be greater than 18 lest we start having negative nickels.  In fact, z cannot be greater than 16 otherwise we break the bank. Actually, if half the coins are nickels, z needs to be much less than 16. Lets start assuming values for z, starting at z=1 and see what happens:

z=1-----------y=52/2=26 now we substitute y=26 and z=1 into eq2:
26=x+1
x=25
CK
26 nickels, 1 dime and 25 pennies---------------half are nickels
and
26*5+1*10+25*1=165
130+10+25=165
165=165

Hope this helps----ptaylor