A collection of nickels, dimes and pennies has an average value of 7¢ per coin. If a nickel were replaced by five pennies, the
average would drop to 6¢ per coin. What is the number of dimes in the collection?
+447 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
