Diane and Michael have a jar containing 41 coins....this is algebra?

Question

this is system of equations.

Diane and Michael have a jar containing 41 coins..All of which are either quarters or nickels. the total value of the coins in the jar is $6.85.

how many of each coin do they have?


please help.

thankyou all!

Answer 1

let

n = nickels

q = quarters

41 = total amount of nickels and quarters

0.05n = value of nickels

0.25q = value of quarters

6.85 = total value of nickels and quarters

- - - - - - - - - - - - -

n + q = 41- - - - - - - - - - - -equation 1
0.05n + 0.25q = 6.85- - - -equation 2
- - - - - - - - - - - - -

Substitute method equation 1

n + q = 41

n + q - n = - n + 41

q = - n + 41

Insert the q value into equation 2
- - - - - - - - - - - - - - - - - - - - - -- - -

0.05n + 0.25q = 6.85

0.05n + 0.25(- n + 41) = 6.84

0.05n + (- 0.25n + 10.25) = 6.85

0.05n - 0.25n + 10.25 = 6.85

- 0.2n + 10.25= 6.85

- 0.2n + 10.25 - 10.25 = 6.86 - 10.25

- 0.2n = - 3.4

- 0.2n / - 0.2 = - 3.4/ - 0.2

n = - 3.4 / - 0.2

n = 17

Insert the n value into equation 1

- - - - - - - - - - - - - -

n + q = 41

17 + q = 41

17 + q - 17 = 41 - 17

q = 24

Insert the q value into equation 1

- - - - - - - - - - - - - - - - -

Check for equation 1

n + q = 41

17 + 24 = 41

41 = 41

- - - - - - - - -

Check for equation 2

0.05n + 0.25q = 6.85

0.05(17) + 0.25(24) = 6.85

0.85 + 6 = 6.85

6.85 = 6.85

- - - - - - - - - - - -

Both equations balance

There are 17 nickels

There are 24 quarters

- - - - - - - -s-

Answer 2

$6.00 in quarters 24 coins
$.85 in nickels 17 coins
-------------
41 coins

Answer 3

Here's your system:

n + q = 41
.05n + .25q = 6.85

-(n + q) = -(41)
20(.05n + .25q) = 20(6.85)

-n + -q = -41
n + 5q = 137

4q = 96

q = 24, n = 17.

Answer 4

0.25x+0.05y=685
x+y=41
25x+5y=685
5x+y=137
-x-y=-41
adding
4x=96
x=24
y=17
so 24 quarters and 17 nickels