You have ten fewer quarters than dimes and 5 fewer nickles than quarters. the total value of the coins is $4.75. how many quarters, nickles, and dimes do you have?

0.05n + 0.10d + 0.25q = 4.75
d = q + 10
n = q - 5

0.05(q - 5) + 0.10(q + 10) + 0.25q = 4.75
0.05q - 0.25 + 0.10q + 1 + 0.25q = 4.75
0.40q + 0.75 = 4.75
0.40q = 4.75 - 0.75
0.40q = 4
q = 4 / 0.40
q = 10 <=== quarters

d = q +10
d = 10 + 10
d = 20 <== dimes

n = q - 5
n = 10 - 5
n = 5 <== nickels

Q=d-10, n=q-5, Solve the first for d and you have:

d=q+10 and n=q-5

.25q+.1d+.05n=4.75, then using d and n from above, this becomes:

.25q+.1(q+10)+.05(q-5)=4.75  perform indicated multiplications

.25q+.1q+1+.05q-.25=4.75  combine like terms

.4q+.75=4.75  subtract .75 from both sides

.4q=4  divide both sides by .4

q=10, since d=q+10 and n=q-5

d=20, n=5

So you have 10 quarters, 20 dimes, and 5 nickels.

It equals 1373.5 that’s the solution