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Mathematics 19 Online
OpenStudy (anonymous):

Solve using a system of equations: There is a collection of nickels and dimes worth $3.30. There are 42 coins in all. How many are there of each kind?

OpenStudy (aripotta):

5n + 10d = 330 n + d = 42

OpenStudy (anonymous):

n is the number of nickels and d is the number of dimes.

OpenStudy (aripotta):

so you gotta use the second equation to solve for d

OpenStudy (aripotta):

or n, but i would do d

OpenStudy (aripotta):

so what does d equal?

OpenStudy (anonymous):

D=576

OpenStudy (aripotta):

no, what you gotta do it find out what d is compared to n. so it's n + d = 42. what you do with that is subtract n from both sides to get d = 42 - n

OpenStudy (aripotta):

and then you can substitute 42 - n for d in the first equation

OpenStudy (aripotta):

so 5n + 10d = 330 would be 5n + 10(42 - n) = 330. is this making sense?

OpenStudy (anonymous):

Yes, but we have to have a system of equations

OpenStudy (aripotta):

the first thing i posted was our system of equations

OpenStudy (aripotta):

now we're solving it

OpenStudy (anonymous):

Oh right sorry

OpenStudy (aripotta):

so do you know what to do from there? from 5n + 10(42 - n) = 330?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so just distribute and then solve from there?

OpenStudy (aripotta):

yea, distribute the 10 into the parentheses and then solve for n and once you know how many nickels there are, just substitute that for n in the second equation and solve for d

OpenStudy (anonymous):

i get a negative

OpenStudy (aripotta):

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