ou find 38 coins consisting only of nickels, dimes, and quarters, with a face value of $4.75. However, the coins all date from 1926, and are worth considerably more than their face value. A coin dealer offers you $8 for each nickel, $6 for each dime, and $22 for each quarter, for a total of $410. How many dimes did you find? Type in your numerical answer only; do not type any words or letters with your answer.

17

Step-by-step explanation:

First of all let us assume the following:

Let the number of nickels = n

Let the number of dimes = d

Let the number of quarters = q

Note also the following conversion factors:

1 nickel = $0.05

1 dime = $0.1

1 quarter = $0.25

we are given the following information:

n + d + q = 38. . . . (1)   (38 coins consisting only of nickels, dimes, and quarters)

0.05n + 0.1d + 0.25q = 4.75 ( we can remove decimals by multiplying through by 100 to give)

5n + 10d + 25q = 475 . . . . (2) note that equation 2 can also be simplified by dividing through by 5, to give the following)

n + 2d + 5q = 95 . . . . . (2)

8n + 6d + 22q = 410 . . . . (3)

Next we will look for a way to eliminate two of the unknowns in each equation to have just one unknown that we can use to solve.

we have three equations

n + d + q = 38. . . . . . . . (1)

n + 2d + 5q = 95 . . . . . (2)

8n + 6d + 22q = 410 . . .(3)

Next we will look for a way to eliminate two of the unknowns in each equation to have just one unknown that we can use to solve.

we can eliminate n by subtracting equation (1) from (2) {eq (2) - eq (1)}

(n - n) + (2d - d) + (5q - q ) = (95 - 38)

= d + 4q = 57 . . . . . . (3)

Next we can multiply eqn (1) through by 8, to make it look like eqn (3), the we can subtract eqn (1) from eqn (3) as follows:

eqn (1) × 8 = 8 × (n + d + q = 38)

8n + 8d + 8q = 304 . . . . (1b) ; subtracting this from eqn (3) we have:

(8n - 8n) + (6d - 8d) + (22q - 8q) = (410 - 304)

= -2d + 14q = 106 . . . . (4)

Now, we have two simplified equations that we can work with:

d + 4q = 57 . . . . . . (3)

-2d + 14q = 106 . . . . (4)

from eqn (3), making 'd' the subject of the formular:

d = 57 - 4q ; replacing the value of d in eqn 4 with this value, we have:

-2(57 - 4q) + 14q = 106

-114 + 8q + 14q = 106

-114 + 22q = 106

22q = 106 + 114 = 220

22q = 220

∴ q = 220 ÷ 22 = 10

∴ q = 10

Next let us solve for 'd' by replacing the value of q with 10 in eqn (3)

d + 4q = 57 . . . . . . (3)    (but q = 10)

d + 4(10) = 57

d + 40 = 57

d = 57 - 40 = 17

d = 17

finally to solve for n, let us input the values of q and d into eqn (1)

n + d + q = 38. . . . (1)  (q = 10 ; d = 17)

n + 17 + 10 = 38

n + 27 = 38

∴ n = 38 - 27 = 11

Therefore the number of each coin are as follows:

nickel = 11

dime = 17

quarters = 10

the solution is quite lengthy, but if you don't understand it at first look, please go through it again.

hi how are you doing today? if you ever need just tell me

step-by-step explanation: