A box contains $8.75 in nickels dimes and quarters there are 51 coins in all and the sum of numbers of nickels and dimes is three less than the number of quarters how many coins of each kind are there?
+2094 Hint: Here are the equations
Let x=nickels
y=dimes
z=quarters.
5x+10y+25z=875
x+y+z=51
x+y+3=z
Hope this helped!
+36915 q + q-3 = 51
q = 27 27 quarters = 6.75 there is 24 coins (n +d) and 2.00 left
.05n + .10 (24-n) = 2.00
-.05n = -.4 n = 8 nickels then 24-n = 16 dimes
+29 simplist method.
5 times 51 is 255
875-255 is 620.
do the rest. iz easy.
