Peter has $4.80 in his piggy bank, consisting of dimes and quarters. if there are 33 coins in all, how many of each does he have?

There would be 10 quarters and 23 dimes.

Step-by-step explanation:

There are generally two methods to solving this type of problem.  (1) is guess and check, where you try different amounts of dimes and quarters in combinations of 33 coins until you reach an amount of $4.80.  However, we can also solve the problem algebraically using two equations and two variables.  In this case, 'd' = amount of dimes and 'q' = amount of quarters.  We know that there are 33 total coins, or d + q  = 33.  We also know that the amount of the coins is $4.80, or .10d + .25q = 4.80.  Solving the first equation for 'q', we get q = 33 - d.  We can then substitute this expression for 'd' in the second equation to solve for d:  .10d + .25(33 - d) = 4.80 or .10d + 8.25 - .25d = 4.80.  Combining like terms and using inverse operations, first subtraction then division, we get d = 23.  If there are 23 dimes, then there must be 10 quarters, for a total of 33 coins or $2.30 in dimes and $2.50 in quarters.

so, converting firstly the mixed fractions to improper fractions and then "adding".