O.k. I'm 42 trying to help my nephew with algebra and lets just say it's been awhile....

Total money $ 101.25
45 more dimes than nickels
3 times as many quarters as dimes.

How many nickels, dimes & quarters?

d times

n nickels
q quarters

d = (n + 45)
q = 3 d so q = 3(n+45)

5 n + 10 d + 25 q = 10125

n + 2 d + 5 q = 2025
so
n + 2 (n+45) + 15 (n+45) = 2025

n + 17(n+45) = 2025
18 n + 765 = 2025
18 n - 1260
n = 70 nickels
you can get the rest now

Got it. Man that felt good to actually solve a problem after all these years.

Hey, at least you are not 74 and trying to help :)

To solve this algebra problem, we need to set up a system of equations based on the given information and then solve for the variables.

Let's define the variables:
Let N = the number of nickels
Let D = the number of dimes
Let Q = the number of quarters

Now let's translate the given information into equations:

1) The total money is $101.25:
The value of each nickel is $0.05, the value of each dime is $0.10, and the value of each quarter is $0.25. The total value of the nickels, dimes, and quarters together is $101.25. So we can write the equation as:
0.05N + 0.10D + 0.25Q = 101.25

2) There are 45 more dimes than nickels:
The number of dimes (D) is equal to the number of nickels (N) plus 45.
D = N + 45

3) There are 3 times as many quarters as dimes:
The number of quarters (Q) is equal to 3 times the number of dimes (D).
Q = 3D

Now we have a system of three equations:
0.05N + 0.10D + 0.25Q = 101.25
D = N + 45
Q = 3D

To find the values of N, D, and Q, we can solve this system of equations. One possible method to solve this is substitution.

Step 1: Solve the second equation (D = N + 45) for N.
N = D - 45

Step 2: Substitute the value of N from Step 1 into the first equation (0.05N + 0.10D + 0.25Q = 101.25).
0.05(D - 45) + 0.10D + 0.25Q = 101.25

Step 3: Simplify and rearrange the equation:
0.05D - 2.25 + 0.10D + 0.25Q = 101.25
0.15D + 0.25Q = 103.50
15D + 25Q = 10350 (multiply both sides by 100 to remove decimals)

Step 4: Now substitute the value of Q (Q = 3D) into the equation from Step 3.
15D + 25(3D) = 10350
15D + 75D = 10350
90D = 10350
D = 115 (divide both sides by 90)

Step 5: Now substitute the value of D into the equation D = N + 45 to find N.
115 = N + 45
N = 115 - 45 = 70

Step 6: Now substitute the values of N and D into the equation Q = 3D to find Q.
Q = 3(115) = 345

So the solution is N = 70, D = 115, and Q = 345. This means there are 70 nickels, 115 dimes, and 345 quarters.