Jerome has some fifty-cent coins and one-dollar coins. He has 3 more one-dollar coins than fifty-cent coins. If he has $13.50, how many one-dollar coins does he have?
To solve this problem, we can use algebraic equations. Let's say Jerome has x fifty-cent coins.
According to the problem, Jerome has 3 more one-dollar coins than fifty-cent coins. So, he has (x + 3) one-dollar coins.
Now, let's calculate the value of Jerome's coins in terms of dollars. The value of x fifty-cent coins is 0.50x dollars, and the value of (x + 3) one-dollar coins is 1 * (x + 3) = (x + 3) dollars.
We know that the total value of Jerome's coins is $13.50. So, we can write the equation:
0.50x + (x + 3) = 13.50
Now, let's solve this equation to find the value of x.
0.50x + x + 3 = 13.50
1.50x + 3 = 13.50
1.50x = 13.50 - 3
1.50x = 10.50
x = 10.50 / 1.50
x = 7
Therefore, Jerome has 7 fifty-cent coins.
Now let's find out how many one-dollar coins he has:
(x + 3) = 7 + 3 = 10
Therefore, Jerome has 10 one-dollar coins.