Question

Joe has a collection of nickels and quarters that is worth $7.40. If the number of nickels were doubled and the number of quarters were increased by 8, the value of the coins would be $10.80. How many quarters does he have?

Answer 1

I've already answered this one today, so you should be able to check my profile for the answer

Answer 2

I just asked the same question......lol

Answer 3

But I didn't understand the answer unfortunately

Answer 4

Let x be the number of nickles, and let y be the number of quarters. Since a nickel is worth 5 cents, the total value of the nickels will be .05 times x, or .05x. Likewise, the total value of the quarters will be .25y. We know the total value of the collection of nickels and quarters is 7.40 so we can write out an equation:
\[.05x+.25y = 7.40\]
Does that make sense so far?

Answer 5

(not done with the problem, just want to make sure we are on the same page)

Answer 6

yes, I got it

Answer 7

Joe has 24 quarters.

Answer 8

alright, now lets look at the next part of the problem. It says the number of nickels was doubled. So if there were x nickels at first, now there will be 2x nickels.
Looking at the quarters, it was increased by 8, so if there were y quarters at first, now there are (y + 8) quarters.

Answer 9

Thank you crazy, but I need to figure out how to set it up and then solve it so that if I have to do it again, I know how instead of coming in here to ask.

Answer 10

I got that part, it was after that I got confused

Answer 11

5p+25q=740
5p(2)+25(q+8)=1080
10p+25q+200=1080
10p+50q=1480

10p+25q=880
25q=600
q=24
p=28
Joe has 24 quarters..

Answer 12

what I nt get is how everyone is going from 10x+25y=800 and getting the answer

Answer 13

Joe has 24 quarters.

Answer 14

crazy go to your own thread

Answer 15

So now using the same idea we did the first time, if each nickel is worth 5 cents, then the value of the nickels will be .05(2x) = x (because .05*2 = 1)
The value of the quarters will be .25(y+8) = .25y+2
So the equation we get is
\[x+.25y+2 = 10.80 \Rightarrow x+.25y = 8.80\]

Answer 16

man deja vu all over again. i could have sworn i just did this

Answer 17

You did, but I got lost and didn't understand your answer

Answer 18

im lost

Answer 19

me too

Answer 20

Now looking at the two equations:
\[.05x+.25y=7.40\]\[x+.25y=8.80\]
we need to find the solution. Noticing that there is a .25y in both equations, I will subtract them to make them cancel out. This is called Elimination. So eq 2 minus eq 1 gives:
\[(1-.05)x = 8.80-7.40 \Rightarrow .95x = 1.40\]

Answer 21

At which part are you lost at?

Answer 22

from where you started the second part of the equation

Answer 23

The\[x+.25y+2 = 10.80\]
part?

Answer 24

yes I dont get how the answer is gotten fromhere

Answer 25

I think for me, that with the answers being so split up into different sections that I can't follow along with what your doing. I have to keep scrolling up to try to figure out what you had and what you did..........

Answer 26

heres what I got so far 5p+25q=740
5p(2)+25(q+8)=1080
10p+25q+200=1080
10p+50q=1480
10p+25q=880
=?

Answer 27

lmao, hi melissa it's diana

Answer 28

lol i figured we were in the same class

Answer 29

So for the second equation, I have twice as many nickels as I did before. So instead of 'x' nickels, i will have 2 times x, or 2x nickels. But each nickel is worth 5 cents. So to figure out how much money I have, I would multiply that 2x by .05, that gives :
\[.05*2x = (.05*2)x = (1)x = x\]
Thats where the x came from.
@lilangl The reason it seems like we are spliting up the information is because these type of problems are called Systems of Equations. We have two things we dont know, the number of quarters and the nickels, so we need two equations to solve this. So once we create the first equation, put it to the side, and create another one.

Answer 30

this site is a godsend

Answer 31

o0o0o i get it now

Answer 32

i still dont get number 4 though

Answer 33

I don't mean that kind of split up, I just meant that part of the answer is about 15 sections up, then the next part is about 10 sections up, then another part 2 sections up............ instead of the answer being all part of the same post so that I can see how it breaks down.

Answer 34

oh oh oh. my apologies on that . That happens a lot >.<

Answer 35

I think I got that one right, let me figure this out and I'll let you know what I got

Answer 36

can i take another crack at this or should i just shut up?

Answer 37

I welcome any and all help satellite

Answer 38

you know your input is valued satellite >.>

Answer 39

well maybe it would just be more confusing than helpful

Answer 40

I'm sorry I'm such an idiot when it comes to this, but generally, once I understand, I can do them by myself

Answer 41

I would appreciate the help....im supposed to be leaving for a party soon-ish lolol

Answer 42

lol, have fun for me Joe, I'll be stuck here trying to figure out how to do my graph problems next

Answer 43

Sure thing :P

Answer 44

already crying just thinking about it

Answer 45

Did you have any luck figuring this out Mel?

Answer 46

honestly i thought i had it but I came close I am trying to use substitution on the equations that joemath was using plus im stuck on number 4 u?

Answer 47

I totally lost it on this one. I got so far and I am lost, and the answers we got here were a little to far spread out to figure out what was going on.............. Did you get my reply on your number 4?