Troubleshooting: A=15-B, C=B+9, D=B+21

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Discussion Overview

The discussion revolves around a system of equations involving four variables, A, B, C, and D, defined by the relationships A=15-B, C=B+9, and D=B+21. Participants are troubleshooting potential mistakes in their calculations related to finding the average of these variables.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant expresses uncertainty about their calculations involving the equations for A, B, C, and D.
  • Another participant presents a system of equations and suggests multiplying each by 3 to simplify the problem.
  • A calculation is provided that leads to an average value of 19.5 for A, B, C, and D.
  • A later reply confirms the previous calculation without further elaboration.

Areas of Agreement / Disagreement

There is some agreement on the calculations leading to the average, but the initial uncertainty about the setup indicates that the discussion remains somewhat unresolved regarding potential mistakes in the original equations.

Contextual Notes

Participants have not fully explored the implications of the relationships between A, B, C, and D, nor have they clarified all assumptions underlying their calculations.

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View attachment 6392 I know that its well above average.

So I got A= 15-B, C=B+9 and D=B+21, but I think I made a mistake somewhere
 

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Let's denote the four numbers by $a$, $b$, $c$ and $d$. Then we have the following system.
\[
\left\{
\begin{aligned}
a+\frac13b+\frac13c+\frac13d=25\\
\frac13a+b+\frac13c+\frac13d=37\\
\frac13a+\frac13b+c+\frac13d=43\\
\frac13a+\frac13b+\frac13c+d=51\\
\end{aligned}
\right.
\]
Multiply each equation by 3 and add them. Recall that you need to find $\dfrac{a+b+c+d}{4}$.
 
6(A+B+C+D)=468

(A+B+C+D)=78

Average is 19.5?
 
You are right.
 

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