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Mathematics 21 Online
OpenStudy (anonymous):

easy pizy! I have 10 cents and 5 cents coins. there are 40 coins in total how many 5 cents are and how many 10 cents do I have in the cup?

OpenStudy (phi):

you need to know how much money (total cents) are in the cup. otherwise, we could say you have 40 nickels and no dimes, for example.

OpenStudy (anonymous):

ok they are aking me to set an equation and from that equation I have to figure out how many coins of 5 cents are and how many coins of 10 cents are.

OpenStudy (phi):

Yes, but they have to tell you how much money the coins add up to.

OpenStudy (anonymous):

Have this individual grab a handful of coins making sure there are only two types of coins in the group (i.e., nickels and dimes, quarters and pennies, pennies and dimes, etc).l your friend should tell you the type of coins they’ve chosen, how many coins they have and the dollar amount of the group. From this information, you will set up two sets of equations and determine how many of each coin they have in their hand.

OpenStudy (anonymous):

this is what they ask me

OpenStudy (phi):

read carefully l your friend should tell you the type of coins they’ve chosen, how many coins they have and the dollar amount of the group.

OpenStudy (anonymous):

I have here nickels and dimes in total there are 40

OpenStudy (phi):

and the dollar amount of the group.

OpenStudy (anonymous):

in dimes there are 1.30$ and in nickels are 1.35

OpenStudy (anonymous):

in total there are 2.65 and here is what I have done so far 0.10+0.052.65 d+n=40c

OpenStudy (phi):

Well, I would not tell the person that, because they could divide 130 by 10 to get 13 dimes and 135/20 to get 27 nickels Just tell how much the dimes and nickels add up to: $2.65 that makes it harder to figure out

OpenStudy (anonymous):

oh ok :)... so what I have done so far is right?

OpenStudy (phi):

0.10+0.052.65<-- this one is not an equation d+n=40c (what is c? just d+n= 40 makes sense)

OpenStudy (anonymous):

amount of coins

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

now there I dont know what else to do

OpenStudy (phi):

if you had d dimes, how much money is that? I would say d*0.10 (the number of dimes times 10 cents per dime) so your first equation should show that the dimes and nickels add up to 2.65

OpenStudy (anonymous):

so lets say 0.10d+0.05n=2.65

OpenStudy (anonymous):

and after that ?? :?

OpenStudy (phi):

yes, you now have 2 equations and 2 unknowns 0.10d+0.05n=2.65 d+n= 40 now you can solve (though you already know the answer!) I would multiply the top equation by 100 to get rid of the decimals (multiply both sides, every term by 100) you get 10d + 5n = 265 d +n = 40 you can do a number of things next. one way is multiply the bottom equation by -5 (every term, both sides) then add the equations, and solve for d

OpenStudy (anonymous):

it gives me 26

OpenStudy (anonymous):

265-5=260 260/10=26

OpenStudy (phi):

no, you are doing something wrong. Start with 10d + 5n = 265 d +n = 40 multiply the bottom equation by -5. what do you get? write down the new equation

OpenStudy (anonymous):

40*-5=200

OpenStudy (anonymous):

-200

OpenStudy (phi):

d +n = 40 write down -5* in front of every term

OpenStudy (anonymous):

d+n=-200

OpenStudy (anonymous):

?

OpenStudy (anonymous):

-5d+5n=-200

OpenStudy (phi):

d +n = 40 write down -5* in front of every term we multiply both sides of the equation (to be fair!) -5*(d+n)= 40* -5 now distribute the -5

OpenStudy (phi):

closer by -5*n is -5n, don't you agree?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so both are negative

OpenStudy (anonymous):

-5d+-5n=-200

OpenStudy (phi):

so you now have 10d + 5n = 265 -5d - 5n = -200 add the equations: write down the left sides added together = the right side added together

OpenStudy (phi):

like this 10d -5d + 5n -5n= 265-200

OpenStudy (phi):

simplify: you have 10 d's and you take away 5 d's. how many d's are left? you have 5 n's and take away 5 n's. how many n's do you have and 265-200 is 65 what do we have for the new equation?

OpenStudy (anonymous):

how did you get265?

OpenStudy (phi):

We had the equation 0.10d+0.05n=2.65 I don't like decimals, so I multiplied both sides of the equation by 100 you get 100*(0.10d + 0.05n)= 2.65*100 distribute the 100: 100*0.10d + 100*0.05n = 265 10d+5n= 265 follow?

OpenStudy (anonymous):

yes

OpenStudy (phi):

so we have 10d + 5n = 265 d +n = 40 then we multiply the bottom equation by -5 (because we want the n's to go away) we get 10d + 5n = 265 -5d -5n = -200 now add the 2 equations. do you know how to do that?

OpenStudy (anonymous):

like @phi said 10d -5d + 5n -5n= 265-200

OpenStudy (phi):

yes, but now simplify it by combining "like terms"

OpenStudy (anonymous):

5d+0n=65?

OpenStudy (phi):

yes. and of course 0*n is 0 and 5d+0 is just 5d so you have 5d= 64 divide both sides by 5 to find d

OpenStudy (phi):

*65 not 64

OpenStudy (anonymous):

correct sorry

OpenStudy (phi):

sorry? I made the typo not you

OpenStudy (anonymous):

64/5=12.8 or 13

OpenStudy (anonymous):

is 65

OpenStudy (phi):

It's 65/5 ( I mistyped)

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

hehe it ok , I made a mistake too

OpenStudy (anonymous):

so is 13

OpenStudy (anonymous):

and for the nickels dont I have to do another equation?

OpenStudy (phi):

yes, now use one of the 2 equations we started with to find n. Choose the easiest one. in this case d+n= 40

OpenStudy (anonymous):

ok

OpenStudy (phi):

replace d with 13: 13+n= 40 add -13 to both sides to find n

OpenStudy (phi):

18?

OpenStudy (anonymous):

wait from 40 sorry

OpenStudy (anonymous):

is going to be n=27

OpenStudy (phi):

adding -13 is the same as subtracting 13. But for me, it is less confusing to always add, and if I want to make something 0 (like +13) I just add -13

OpenStudy (phi):

yes d= 13 and n=27, which are the numbers you picked in the first place. I hope all this make sense.

OpenStudy (anonymous):

hehe yes it does ...Thank you so so so much for helping me! ;)

OpenStudy (phi):

If you want more info, here is a video that is helpful http://www.khanacademy.org/math/algebra/systems-of-eq-and-ineq/v/solving-systems-by-elimination

OpenStudy (anonymous):

Ok thank you1, I will watch that video :)

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