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# Algebra Advanced systems of equations

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Amy, Ben, Carl, and Debbie each have some coins. Ben has three times the number of coins that Amy has and a third of the number of coins that Carl has, and Debbie has two-thirds the number of coins that Carl has. The number of coins that Amy has, multiplied by the number of coins that Ben has, multiplied by the number of coins that Carl has, multiplied by the number of coins that Debbie has, is 162. How many coins do the four children have all together?

That is my question.

Sep 2, 2020

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Good job ilorty.

Sep 2, 2020
edited by SHADOWwolf  Sep 2, 2020
#2
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@ SHADOWwolf, don't you see that the link brings you back to the same page? What is the point of the link if it only brings you to the same page? Who are you trying to help? Are you just a troll that asks questions and clogs up the forum with "not so useful" links and answers like my hair clogs the bathroom drain? Or is it the wrong link? (The preview is there for a reason)

Anyways, for your question we will need variables:

A = Amy's coins

B = Ben's coins

C= Carl's coins

D= Debbie's coins

With what we know, we can say this:

B = 3A

3B = C

3D = 2C

ABCD = 162

First and foremost, we need to write each variable in terms of one variable. Let's use A:

A = A

B = 3A

C = 3B = 3(3A) = 9A

D = 2/3C = 6A

Therefore:

ABCD = A (3A) (9A) (6A) = 162A^4 = 162

A = 1

Now that we know that A is 1, we can say that:

A = 1

B = 3(1) = 3

C = 9(1) = 9

D = 6(1) = 6

To check the work, let's multiply:

1 * 3 * 9 * 6 = 162.

:)

Sep 2, 2020
#3
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When I use the link it takes me to a page where the question has already been answered.  Sorry if you can't see it.

Sep 2, 2020
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I think it is the wrong link... it leads to this exact page!

ilorty  Sep 2, 2020
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Amy # = A

B = 3A

C =9A

D = 6A             (this is all given in the Q)

A * 3A * 9A * 6A = 162

A = 1

then

B = 3

C = 9

D = 6

Sep 2, 2020