A collection of nickels, dimes and pennies has an average value of 7 cents per coin. If a nickel were replaced by five pennies, the average

$\dfrac{p + 5n + 10d}{p+n+d}=7 \\ \dfrac{p+5 + 5(n-1) + 10d}{p+5+(n-1)+d} = \dfrac{p+5n+10d}{p+n+d+4}=6 \\ 7(p+n+d) = 6(p+n+d+4) \\ p+n+d = 24$

$p+5n+10d = (24)(7) = 168 \\ \text{clearly }p=3 \text{ so }\\ 5n+10d = 165,~n+d = 21 \\ 5(21-d)+10d = 165 \\ 105 + 5d = 165 \\ 5d = 60 \\ d=12 \\ n=21-12 = 9$

$p=3,~n=9,~d=12$

Solve the following system:
{3 + 10 d + 5 n = 168 | (equation 1)
5 + 10 d + 5 (n - 1) + p = 168 | (equation 2)
d + n + p = 24 | (equation 3)

Express the system in standard form:
{5 n + 10 d+0 p = 165 | (equation 1)
5 n + 10 d + p = 168 | (equation 2)
n + d + p = 24 | (equation 3)

Subtract equation 1 from equation 2:
{5 n + 10 d+0 p = 165 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
n + d + p = 24 | (equation 3)

Divide equation 1 by 5:
{n + 2 d+0 p = 33 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
n + d + p = 24 | (equation 3)

Subtract equation 1 from equation 3:
{n + 2 d+0 p = 33 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
0 n - d + p = -9 | (equation 3)

Swap equation 2 with equation 3:
{n + 2 d+0 p = 33 | (equation 1)
0 n - d + p = -9 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Subtract equation 3 from equation 2:
{n + 2 d+0 p = 33 | (equation 1)
0 n - d+0 p = -12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Multiply equation 2 by -1:
{n + 2 d+0 p = 33 | (equation 1)
0 n+d+0 p = 12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{n+0 d+0 p = 9 | (equation 1)
0 n+d+0 p = 12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

n = 9 Nickels
d = 12 Dimes
p = 3 Pennies