algebra word problem

Question

I have 30 coins, all nickels, dimes, and quarters, worth $4.60. There are two more dimes than quarters. How many of each coin do I have.

Answer 1

We have 3 unknowns, so we need 3 equations to solve this.

Let x = # of nickels

y = # of dimes

z = # of quarters

Eq. 1: x + y + z = 30 The number of coins adds up to 30

Eq. 2: y = z + 2 There are 2 more dimes than quarters

Eq. 3: .05x + .10y + .25z = 4.60 Total amount is equal to $4.60

Using the value of Eq. 2 for "y", we can substitute this into Eq. 1 and solve for "z":

x + y +z = 30 Eq. 1

x + (z + 2) + z = 30 substitute for "y"

x + 2z + 2 = 30 do addition of the "z's

x = 30 - 2 - 2z subtract 2 and 2z from both sides

x = 28 - 2z simplify

Now, put these value for "x" and the value for "y" given into Eq. 3 and solve for "z":

.05x + .10y + .25z = 4.60 Eq. 3

.05(28 - 2z) + .10(z + 2) + .25z = 4.60 substitute for "x" and "y"

1.4 - .10z + .10z + .20 + .25z = 4.60 do multiplication

1.60 + .25z = 4.60 simplify

.25z = 4.60 - 1.60 subtract 1.60 from both sides

.25z = 3.00 simplify

z = 12 divide both sides by .25

Using this value for "z" in Eq. 2 we can solve for "y":

y = z + 2 Eq. 2

y = 12 + 2 substitute for "z"

y = 14

Using these values for "y" and "z" in Eq. 1 we can solve for "x":

x + y + z = 30 Eq. 1

x + 14 + 12 = 30 substitute for "y" and "z"

x + 26 = 30 simplify

x = 30 - 26 subtract 26 from both sides

x = 4

Therefore, the numbers are: nickels = 4

dimes = 14

quarters = 12

Hope this helps!

Answer 2

You can solve this problem by setting up a system of linear equations. Let n be the number of nickels, d be the number of dimes, and q be the number of quarters. Then, we have:

5n + 10d + 25q = 460

n + d + q = 30

d - q = 2

We want to eliminate n from our system. Subtracting 5 times the second equation from the first equation yields:

5d + 20q = 310

d - q = 2

We can eliminate either of the remaining variables, so let's eliminate d. Subtracting 5 times the second equation from the first equation yields:

25q = 300

q = 12

Then, plugging back into the third original equation, we have:

d - 12 = 2

d = 14

And we plug both back into the second original equation to get:

n + 12 + 14 = 30

n = 4

Therefore, we have 4 nickels, 14 dimes, and 12 quarters.