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