J has 16 dimes, qtrs; W has twice dimes, 1/3 qtrs. If....

sandyk109

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Jessica has 16 dimes and quarters. Whitney has twice twice as many dimes and 1/3 as many quarters as Jessica has. If they both have the same amount of money, what coins does each have?

10(d-16) + 24q = 2(10d) +1/3 25q
10d -160 +25q =20d +1/3 25q
 
Re: Beginning Algebra

Jessica has 16 dimes and quarters. Whitney has twice twice as many dimes and 1/3 as many quarters as Jessica has. If they both have the same amount of money, what coins does each have?

10(d-16) + 24q = 2(10d) +1/3 25q
10d -160 +25q =20d +1/3 25q

I'm not quite sure what you are saying. And you don't say what method you are supposed to use. I'll assumeyou are to use two equations in two unknowns. Possibly you are not recognizing that the number of coins and the value of the coins (amount) are two different things. That's why it is so important to name things to get started. For instance...

Let d stand for the number of dimes that Jessica has.
Let q stand for the number of quarters that Jessica has.

Now you can say that the value of the dimes that Jessica has is 10d and the value of the quarters that Jessica has is 25q.

You are also ready to say that the number of dimes that Whitney has is 2d and the number of quarters that Jessica has is q/3.

Furthermore, the amount (value) of Jessica's money is 10d + 25q.
And the amount of Whitney's money is 10(2d) + 25(q/3).

Now that you have everything properly named, you should be ready to build your equations and solve simultaneously.
 
Re: Beginning Algebra

sandyk109 said:
Jessica has 16 dimes and quarters. Whitney has twice twice as many dimes and 1/3 as many quarters as Jessica has. If they both have the same amount of money, what coins does each have?

Let's try it using one unknown and see what happens.

First, define what you know.

\(\displaystyle Let \ \ d = no.\ \ dimes \ \ Jessica \ \ has\)

\(\displaystyle Let \ \ 16-d = no.\ \ quarters \ \ Jessica \ \ has\)

\(\displaystyle Let \ \ 2d = no.\ \ dimes \ \ Whitney \ \ has\)

\(\displaystyle Let \ \ \frac{1}{3}(16-d) = no.\ \ quarters \ \ Whitney \ \ has\)

All that came from the first two sentences.

A dime is worth 10 cents. So d dimes would be worth 10d cents, and 2d dimes would be worth 10(2d) cents.

Similarly, a quarter is worth 25 cents, so 16-d quarters would be worth 25(16-d) cents, and \(\displaystyle \frac{1}{3}(16-d)\) quarters would be worth \(\displaystyle 25(\frac{1}{3})(16-d)\) cents.

Since Jessica and Whitney have the same amount of money, we can build this equation:

\(\displaystyle \overbrace{10d+25(16-d)}^{\text{Jessica's amount}}=\overbrace{10(2d)+25(\frac{1}{3})(16-d)}^{\text{Whitney's amount}}\)

I would probably multiply each term by 3 to eliminate the fraction and then solve from there.

Can you finish from here? Once you solve for d, substitute that back into the definitions we started with to find out how many dimes and quarters each girl has.

Good luck. Let us know if you were successful.
 
Re: Beginning Algebra

Another way to skin that cat:

Jessica: x dimes and 3y quarters ; then:
Whitney: 2x dimes and y quarters

10x + 25(3y) = 10(2x) + 25y
x = 5y

SO Jessica has 5y + 3y coins; 8y = 16 (given)
Finish it.
 
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