Question 23736
Let x = no of nickels
x+4 = no of dimes

The total of the nickels and dimes is 2x + 4, and the total number of coins is 25.  Therefore the number of quarters is actually 25 - (total of nickels and dimes).


25-(2x+4) = no of quarters
25 - 2x - 4 = no quarters
21 - 2x = no quarters


Values of the coins (in cents) is the value of each coin (in cents) times the number of coins:
5*x = value of nickels
10*(x+4)= value of dimes
25*(21-2x) = value of quarters


Equation is the total value of the coins in CENTS = 320:

5x + 10(x+4) + 25(21-2x) = 320
5x + 10x + 40 + 525 - 50x = 320
-35x + 565 = 320


Subtract 565 from each side:

-35x + 565 - 565 = 320-565
-35x = -245


Divide both sides by -35 and hope it comes out even!!
{{{-35x = -245}}}
{{{(-35x)/-35 = -245-35 }}}
{{{x= 7}}} Nickels
{{{x+4 = 11}}} Dimes
{{{21-2x = 7 }}} Quarters


Check the Values:
x = 7 Nickels = $0.35
x+4 = 11 Dimes= $1.10
21-2x = 7 Quarters = 1.75
TOTAL Value of coins = $3.20
It checks!!


R^2 at SCC