A box contains 83 coins, only dimes and nickels. The amount of money in the box is $5.80. How many dimes and how many nickels are in the box?

Step-by-step explanation:

n = nickels

d = dimes

q = quarters

1)  n + d + q = 39

n+d = q - 3

Translate to:  2)   n + d - q = -3

.05n + .1d + .25q = 6.7   amount of each coin times the number of coins must add to 6.7

Multiply both sides by 20...

3)  1n + 2d + 5q = 134

Combine 1) & 2)

n + d + q = 39

n + d  - q = -3

Add together

2n + 2d = 36

Divide by 2:

4)  n + d = 18

Combine 1) and 3)

n + d + q = 39         Multiply by -5 =>  -5n -5d -5q = -195

n + 2d + 5q = 134                              n + 2d + 5q = 134

Add together->                                 5)   -4n - 3d  = -61

Combine 4) and 5)

  n + d = 18   Multiply by 4 =>  4n + 4d = 72

-4n - 3d = 61                          -4n - 3d = -61

Add:                                               d = 11

11 dimes...

Plug that into 4)

n + 11 = 18

n = 18-11 = 7 nickels.

Plug that into 1.

n + d + q = 39

7 + 11 + q = 39

18 + q = 39

q = 39-18 = 21 quarters

21 quarters, 7 nickels, and 11 dimes.

21(.25) + 7(.05) + 11(.1) = 5.25 + .35 + 1.1 = 5.6 + 1.1 = 6.7   Check.

54

step-by-step explanation:

6x9

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.