MATH SOLVE

4 months ago

Q:
# You have 23 coins, including nickels, dimes, and quarters. if you have two more dimes than quarters, and the total value of the coins is $2.50. how many of each kind of coin do you have?

Accepted Solution

A:

You can write three equations in the numbers of nickels (n), dime (d), and quarters (q).

n + d + q = 23 . . . . . . . there are 23 coins total

0n +d -q = 2 . . . . . . . . .there are 2 more dimes than quarters

5n +10d +25q = 250 . .the total value is $2.50

The collection includes 11 nickels, 7 dimes, and 5 quarters.

_____

I used the matrix function of my calculator to solve these equations. You can find q by subtracting from the last equation five times the sum of the first two equations.

(5n +10d +25q) -5((n +d +q) +(d -q)) = (250) -5(23 +2)

25q = 125 . . . . . . . simplify

q = 5

From the second equation,

d = q +2 = 7

And from the first,

n = 23 -5 -7 = 11

n + d + q = 23 . . . . . . . there are 23 coins total

0n +d -q = 2 . . . . . . . . .there are 2 more dimes than quarters

5n +10d +25q = 250 . .the total value is $2.50

The collection includes 11 nickels, 7 dimes, and 5 quarters.

_____

I used the matrix function of my calculator to solve these equations. You can find q by subtracting from the last equation five times the sum of the first two equations.

(5n +10d +25q) -5((n +d +q) +(d -q)) = (250) -5(23 +2)

25q = 125 . . . . . . . simplify

q = 5

From the second equation,

d = q +2 = 7

And from the first,

n = 23 -5 -7 = 11