Trip has 15 coins worth 95 cents. four of the coins are each worth twice as much as the rest. construct a math argument to justify the conjecture that trip has 11 nickles and 4 dimes.

Let's assign x to the value of one coin. Since four coins are twice the value of x, we'll have the term  or  for these coins. The rest of the eleven coins have the value of x thus we'll have  as one of the terms, too.

If we add these two terms, we should get a sum of $0.95 since that's the total worth of Trip's coins. We can then solve for x to find how much each of the 11 coins of Trip are worth.





Since 11 coins of Trip are worth $0.05 each, we have successfully verified that Trip has 11 nickles. Furthermore, the 4 remaining coins would automatically be dimes since it's twice the value of nickel.