freida has 35 coins valued at $7.15. if she only has nickels and quarters, how many nickels does she have

Question

not sure how to solve

Answer 1

We solve this by treating it as a pair of linear equations:

n + q = 35 (she has 35 nickels and quarters all together)

.05n + .25q = 7.15 (the total change of the coins is $7.15)

 

Now like any linear pair, we can use the substitution or elimination methods. I think I'll go with elimination in this case. By multiplying the second equation by -4, we'll have a negative one q:

  n + q  =  35

-.2n - q = -28.6

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

.8n = 6.4

 

Divide both sides by .8 to isolate n:

n = 8

 

Now we can solve for q:

8 + q = 35

q = 35 - 8

q = 27

 

Now to check:

.05 * 8 + 27 * .25 = 7.15

.40 + 6.75 = 7.15

7.15 = 7.15

Answer 2

Let's write down what we know by using algebra.

 

I'm going to use the variables Q and N for quarters and nickels, so that I can keep track of them all. If Q is how many quarters I have and N is how many nickels I have, and I have 35 coins total, then I know:

 

Q + N = 35

 

I also know how much quarters and nickels are worth. If I multiply the number of quarters by .25 and I multiply the number of nickels by .05, then I should get the amount the coins are worth. I can use that to make this equation:

 

0.25Q + 0.05N = 7.15

 

Now I have two equations, so I can make a system of equations and solve them using elimination by multiplication

 

Q+N=35

0.25Q+0.05N=7.15

 

Q+N=35

Q+0.2N=28.6

 

0Q+0.8N=6.4

 

N=8

 

N=8 is saying there are eight nickels, but you should probably double check that this number makes sense.

 

Let me know if you have any questions!