Question

Joe has a collection of nickels and dimes that is worth $6.10. If the number of dimes was tripled and the number of nickels was increased by 49, the value of the coins would be $18.95. How many nickels and dimes does he have?

Answer 1

What is 49 times the value of a nickel?

Answer 2

^Yeah, I'm not too sure of the values either.

Answer 3

i dont know im so confused

Answer 4

If you figure out how much those 49 nickels are worth and add them to the original 6.10 then you can take the rest of the money and see how many dimes it is.

Answer 5

It's 49 more nickels, not 49 times as many

Answer 6

You have to multiply the 49 by .o5 because then you know how much the 49 nickels value @JakeV8

Answer 7

ok..

Answer 8

Confirmation on what @JakeV8 said. That 2nd equation should be (49 + x)(0.05) + 3y(0.10) = 18.95

Answer 9

wait for @tcarroll010's answer... it will help :)

Answer 10

you guys are confusing me haha

Answer 11

After you get the value of the nickels, x, you add it to the original 6.10. What do you get? @jdorta1

Answer 12

Just take Mathmuse's first equation and the second equation I gave, which JakeV8 actually gave.

Answer 13

@tcarroll010 so its x(0.05)+y(0.10)=6.10 and (49+x)(0.05)+3y(0.10)=18.95?

Answer 14

Yes, but credit goes to Mathmuse and JakeV8.

Answer 15

ok so what do i do from there? i am seriously horrible at this stuff

Answer 16

It might sound simplistic, but I'd get rid of those decimal points first by multiplying each side of both equations by 100.

Answer 17

ok so it would be x(5)+y(10)=6.10?

Answer 18

Perhaps you should listen to @Mathmuse or @JakeV8 at this point, because whether you listen to me or them, I don't deserve the credit here, but I'll stick around to help. I suggest you give one or the other a medal when we're done and then I'll medal the other guy. And that's 610 on the right side.

Answer 19

OK

Answer 20

I am having trouble keeping the connection going here... sorry... my page froze for the last 2 minutes or so.

I don't mind helping either, and I'm not worried about the credit. I just hate making a comment and then having the page freeze, leaving everyone waiting :(

I was going to medal tcarroll010 for helping :) Glad we're all friendly here :)

Answer 21

@jdorta1, when you multiply that first equation through by 100, you need to be sure to multiply both sides... so the right side goes from 6.10 to 100*6.10 or 610.

Answer 22

OK I GOT THAT

Answer 23

so i now have x(5)+y(10)=610

Answer 24

opps sorry for caps lol

Answer 25

5(49 + x) + 30y = 1895 for the other equation. With this equation, expand 5(49 + x)

Answer 26

ok

Answer 27

@jdorta1, What did you get for the 2nd equation after you expanded the stuff on the left like @tcarroll010 suggested?

Answer 28

@JakeV8 oh my gosh i feel so dumb.. i have no idea what your talking about..

Answer 29

Equation 1: x(5)+y(10)=610
Equation 2: 5(49 + x) + 30y = 1895

Now simplify equation 2 by multiplying the 5 across the parenthesis... expand by multiplying 5(49 + x). Then rewrite both Equation 1 and the new simplified version of Equation 2.

Answer 30

@JakeV8 so it would be 245+5x+30y=1895?

Answer 31

yes, good. Now, simplify more by moving the 245 from the left to the right of the equation... you have to subtract 245 from both sides. Then re-write the updated version of Equation 2.

Answer 32

5x+30y=1651?

Answer 33

check that subtraction again....

Answer 34

oh shoot my bad haha its 1650

Answer 35

ok, so you should have two equations with the x's and y's on the left, and the numbers on the right... that's the form you need to aim for when you're setting this up.

Equation 1: 5x + 10y = 610
Equation 2: 5x + 30y = 1650

Answer 36

Now you can finally get started on solving for x and y.

Answer 37

@JakeV8 yes i have that

Answer 38

@JakeV8 ok...how? lol

Answer 39

@JakeV8 dont i subtract y from both sides or something

Answer 40

Well, do you see how there is a 5x in both equations? The best technique here is called "elimination"....

Answer 41

@JakeV8 ok i see that..

Answer 42

Equation 1: 5x + 10y = 610
Equation 2: 5x + 30y = 1650

Let's subtract the first equation from the second. When you subtract, since "5x" appears in each, it will be eliminated.

Eq 2: 5x + 30y = 1650
- Eq 1: 5x + 10y = 610
-------------------------

Answer 43

@jakeV8 So its now 20y=1040

Answer 44

@jakeV8 then i divide both sides by 20 correct?

Answer 45

yay! Good thinking :)

Answer 46

@JakeV8 y=52

Answer 47

Correct! So, that means you've solved for "y". How will you solve for "x" now?

Answer 48

@JakeV8 plug in y

5x+30(52)=1650

Answer 49

You're doing well... continue :)

Answer 50

5x+1560=1650
5x=90
x=18

Answer 51

Correct again :)

So, here's the place where you make sure you get this right... terrible to do all that work and then miss it... you need to double check your solutions in the original problem to make sure everything works.

Joe has a collection of nickels and dimes that is worth $6.10. If the number of dimes was tripled and the number of nickels was increased by 49, the value of the coins would be $18.95. How many nickels and dimes does he have?

You found x = 18 and y = 52.

So you should use those numbers and see if they work in the word problem.

Answer 52

@JakeV8 ok and how do you do that?

Answer 53

18 nickels: 0.05 * 18 = $0.90
52 dimes: 0.10 * 52 = $5.20
------------------------------
$6.10 <---- works so far...

Answer 54

ooook I see where your going...

Answer 55

The other statement said that if you tripled your dimes and had 49 nickels, the new value would be $18.95.

76
JakeV8
Medals 1

49 + 18 nickels = 67 nickels: 0.05 * 67 = $3.35
3 * 52 dimes = 156 dimes: 0.10 * 156 = $15.60
------ --------------------------------------
$18.95 (this also works)

So it's correct!! 18 nickels and 52 dimes.

Answer 56

oops, some weird copy paste thing happened there with the medals stuff ??

Answer 57

YAY!!! thank you so much you are the greatest!!! <3

Answer 58

I think I must have copied my name and score or something... weird. oh well.

Do you follow the "double-check" process? If it doesn't all work out, you know you made an error somewhere earlier. But if it does all work in both equations, then you are 100% guaranteed to get it correct :)

Answer 59

you are awesome! lol

Answer 60

Glad to help... :)

once you are comfortable solving a system of 2 equations like this, they're all pretty similar. The other tricky thing is to learn to "translate" the word problem into math-language... you have to get those words into a form where it's 2 equations with 2 unknowns. The big clue is that they ask for "how many nickels and how many dimes?" so you know you have 2 unknowns... therefore you need 2 equations from somewhere...

Good luck!

Answer 61

I really do appreciate the help =)