Word problem: System of equations 4 variables

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I have been stumped for over a week please help. :cry:
There are four boxes labeled A, B, C, and D. A is 3 times as heavy as B, B is 3 times as heavy as C, and C is 3 times as heavy as D. The weight of D is 12 times less than the combined weight of B and C. What are the weights of the four boxes?
 
Brian G said:
I have been stumped for over a week
Click to expand...
Please reply showing the system of equations you have set up. Specify the method you are using for solving the system (Cramer's Rule? matrix row operations? Gauss-Jordan elimination? etc), and show how far you have gotten and where you are stuck.

Thank you.

Eliz.
 
Brian G said:
I have been stumped for over a week please help. :cry:
There are four boxes labeled A, B, C, and D. A is 3 times as heavy as B, B is 3 times as heavy as C, and C is 3 times as heavy as D. The weight of D is 12 times less than the combined weight of B and C. What are the weights of the four boxes?
Click to expand...
No 4 variables required there, Brian; let x = D's weight; then:
D=x, C=3x, B=9x, A=27x
The fact that "weight of D is 12 times less than B+C" is evident,
since B+C = 12x: so that's no clue.
So weights can be anything you want...any value can be assigned to x.
CHECK the problem...something missing for sure :shock:
 
I totally do not know where to start. This is the problem as it was given without instruction. All that was said is it was an Algebra 2 problem. :?
 
Brian G said:
I totally do not know where to start.
Click to expand...
Try starting from the step-by-step instructions, provided earlier. If you get stuck, please reply showing how far you have gotten.

Thank you.

Eliz.
 
Brian G said:
I have been stumped for over a week please help. :cry:
There are four boxes labeled A, B, C, and D. A is 3 times as heavy as B, B is 3 times as heavy as C, and C is 3 times as heavy as D. The weight of D is 12 times less than the combined weight of B and C. What are the weights of the four boxes?
Click to expand...
How about:

A = 3B
B = 3C
C = 3D
D = B + C - 12
:
A few substututions will take you to your answer.
 
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