Answer:
You have 15 dimes. You have 9 quarters. You have 12 nickels
Step-by-step explanation:
lets set some variables:
let "n" = the number of nickels
let "d" = the number of dimes
let "q" = the number of quarters
So, the total amount of money you have should be: $4.35 = 0.25q + 0.10d + 0.05n
Now let's look at the relationships between the coins:
"There are three fewer quarters than nickels": n - 3 = q
"six more dimes than quarters": q + 6 = d
So now you have three equations with three variables, all you need to do is solve.
![\left \{ {{4.35 = 0.25q + 0.10d + 0.05n} \atop {n - 3 = q}} \atop {q + 6 = d}}\right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B4.35%20%3D%200.25q%20%2B%200.10d%20%2B%200.05n%7D%20%5Catop%20%7Bn%20-%203%20%3D%20q%7D%7D%20%5Catop%20%7Bq%20%2B%206%20%3D%20d%7D%7D%5Cright.)
first, you can substitute "n-3" for "q" (according to the 2nd equation) in the 1st and 3rd equation, you get:
![\left \{ {{4.35=0.25(n-3)+0.10d+0.05n} \atop {(n-3)+6=d}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B4.35%3D0.25%28n-3%29%2B0.10d%2B0.05n%7D%20%5Catop%20%7B%28n-3%29%2B6%3Dd%7D%7D%20%5Cright.)
You now only have two equations and two variables.
Simplify:
![\left \{ {{4.35=0.25n-0.75+0.10d+0.05n} \atop {n+3=d}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B4.35%3D0.25n-0.75%2B0.10d%2B0.05n%7D%20%5Catop%20%7Bn%2B3%3Dd%7D%7D%20%5Cright.)
![\left \{ {{4.35=0.30n-0.75+0.10d} \atop {n+3=d}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B4.35%3D0.30n-0.75%2B0.10d%7D%20%5Catop%20%7Bn%2B3%3Dd%7D%7D%20%5Cright.)
Now substitute "n+3" for "d" (according to the 2nd equation) in the 1st equation:
4.35=0.30n-0.75+0.10(n+3)
simplify:
4.35=0.30n-0.75+0.10n+0.30
4.35=0.40n-0.45
4.35+0.45=0.40n
4.80=0.40n
n=12
You have 12 nickels. Now sub "n" back into your equations to find the number of dimes and quarters:
n - 3 = q
12 - 3 = q
q = 9
You have 9 quarters.
q + 6 = d
9 + 6 = d
d = 15
You have 15 dimes.