Question 392072
A collection of 90 coins, pennies, nickels, and dimes, has a value of $2.85.
 If there are twice as may pennies as there are nickels and dimes combined, how many pennies, dimes, and nickels are there?
:
p + n + d = 90
:
.01p + .05n + .10d = 2.85
:
"there are twice as may pennies as there are nickels and dimes combined,"
p = 2(n + d)
divide both sides by 2
.5p = n + d
.5p - n - d = 0; add to the 1st equation
  p + n + d = 90
----------------------eliminates n and d find p
1.5p = 90
p = {{{90/1.5}}}
p = 60 pennies
:
Rewrite the two equation using p=60
n + d = 30
n = (30-d)
and
.05n + .10d = 2.25
Substitute (30-d) for n
.05(30-d) + .10d = 2.25
1.5 - .05d + .10d = 2.25
.05d = 2.25 - 1.50
d = {{{.75/.05}}}
d = 15 dimes
then obviously
n = 15 nickels
:
:
Check solutions by finding the total$
.01(60) + .05(15) + .10(15) =
.60 + .75 + 1.5 = 2.85, confirms our solution p=60; n=15; d=15