For problems like this we must take what we know and make a system of equations which is a fancy name for a group of equations that can be used to solve one another.
The first equation we can make is the one which takes into account the total amount we have: 10D + 5N + P = 110 (where D is the number of dimes, N is the number of nickels and P is the number of pennies.
Because we have three variables, we need three equations to solve for each one. The problem states that we have three times as many nickels as pennies so we can write the equation N = 3P.
The final equation we can make is from the statement saying there are two less dimes than pennies, we can write this as D+2=P
Now we must use all these equations to solve for a single variable.
D+2=P
N=3P
10D + 5N + P = 110
Lets solve for P by changing everything to Ps by substituting the top two equations into the bottom one to get:
10(P-2) + 5(3P) + P = 110
10P - 20 +15P + P = 110
26P = 130
P=5
We can plug this into the smaller equations to find
N=3P=3(5)=15 and D+2=P, D+2=5, D=3
Dimes: 3
Nickels: 15
Pennies: 5