Question 168398
Let x=number of nickels; 2x=twice the number of nickels; 2x-8 would be 8 fewer than twice the number of nickels
y=number of dimes; 2y=twice the number of dimes; 2y+2=2 more than twice the number of dimes
z=number of quarters

Now lets deal in pennies instead of $.
Value of the nickels=5x
Value of the dimes=10y
4 times the value of the nickels and dimes together=4(5x+10y)
Value of the quarters=25z

Now we are told the following:
x=2y+2 -------------------eq1
z=2x-8 ---------------------eq2
and
25z=4(5x+10y)+160-------------------------eq3

Lets rearrange and simplify the above equations so we can see what we are dealing with:
x-2y   =2---------------------eq1
2x   -z=8---------------------eq2
4x+8y-5z=-32-------------------eq3
Multiply eq1 by 4 and add it to 
eq3 and we get:
8x-5z=-24----------------------------eq3a
Next multiply eq2 by 4 and subtract it from eq3a and we get

(8x-5z+-24) -(8x-4z=32) or
-z=-56
z=56-------------------------------------------number of quarters
substitute z=56 into eq2 and we get:
2x-56=8
2x=64
x=32-------------------------------------number of nickels
substitute x=32 into eq1 and we get:
32-2y=2
-2y=-30
y=15------------------------------------- number of dimes
CK
two more nickels than twice the number of dimes: 
2*15+2=32
32=32
eight fewer quarters than twice the number of nickels:
32*2-8=56
56=56
value of the quarters was $1.60 greater than four times the value of the nickels and dimes together:
56*25=4(32*5+15*10)+160
1400=4(160+150)+160=(4*310)+160
1400=1240+160=1400
1400=1400

Hope this helps-----ptaylor