Question 609354
Let p = pennies
Let n = nickels
Let d = dimes

Twice as many nickels as pennies:
n = 2p

Ten more dimes than nickels:
d = n + 10

165 coins altogether:
p + n + d = 165

Now that you have 3 equations and 3 variables, you can solve:
n = 2p
d = n + 10
p + n + d = 165

Substitute the n value into the other two equations:
d = 2p + 10
p + 2p + d = 165  ... which simplifies to ---> 3p + d = 165
 

You have reduced the number of equations and variables to 2. Keep going:

Substitute d into the second equation:
3p + 2p + 10 = 165
5p + 10 = 165
5p = 155
p = 31

Then plug p back into the other equation
d = 2(31) + 10
d = 62 + 10
d = 72

Then plug p and d back into one of the original equations:
p + n + d = 165
31 + n + 72 = 165
103 + n = 165
n = 62

Check:
n = 2p
62 = 2(31)

d = n + 10
72 = 62 + 10

p + n + d = 165
31 + 62 + 72 = 165

You're good to go. ^_^