To solve the problem of how many dimes and quarters Joette has, we will use algebra to set up a system of equations and then solve the system step-by-step.
1. Define the variables:
- Let \(\(d\)\) be the number of dimes.
- Let \(\(q\)\) be the number of quarters.
2. Set up the equations:
- We know that the total value of the coins is \($16.95.
- Each dime is worth $\)0.10, so the total value contributed by the dimes is \(\(0.10d\)\).
- Each quarter is worth $0.25, so the total value contributed by the quarters is \(\(0.25q\)\).
- This gives us the first equation:
\(\[
0.10d + 0.25q = 16.95
\]\)
- The problem also states that the number of quarters is three more than twice the number of dimes. This gives us the second equation:
\(\[
q = 2d + 3
\]\)
3. Substitute the second equation into the first:
- We can replace \(\(q\)\) in the first equation with \(\(2d + 3\)\):
\(\[
0.10d + 0.25(2d + 3) = 16.95
\]\)
4. Simplify and solve for \(\(d\)\):
- Expand the equation:
\(\[
0.10d + 0.25(2d) + 0.25(3) = 16.95
\]\)
\(\[
0.10d + 0.50d + 0.75 = 16.95
\]\)
- Combine like terms:
\(\[
0.60d + 0.75 = 16.95
\]\)
- Subtract 0.75 from both sides to isolate terms involving \(\(d\)\):
\(\[
0.60d = 16.20
\]\)
- Divide both sides by 0.60 to solve for \(\(d\)\):
\(\[
d = \frac{16.20}{0.60} = 27
\]\)
- So, the number of dimes \(\(d\)\) is 27.
5. Find the number of quarters \(\(q\)\) using the second equation:
- Substitute \(\(d = 27\)\) into the equation \(\(q = 2d + 3\)\):
\(\[
q = 2(27) + 3 = 54 + 3 = 57
\]\)
- So, the number of quarters \(\(q\)\) is 57.
6. Conclusion:
- Joette has 27 dimes and 57 quarters.
