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$$(x + y)/2 = 400$$ $$(y + z)/2 = 800$$

I need to know what is $$(x+y+z)/3$$

nmasanta
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roglol
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  • Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. Also, please format you questions with MathJax – saulspatz Jul 18 '19 at 15:51
  • Do you mean $$\frac{x+y}{2}=400\frac{y+z}{2}=800$$? – Dr. Sonnhard Graubner Jul 18 '19 at 15:52
  • Why not try some examples? Can you come up with some triples $(x,y,z)$ that satisfy your first two equations? Do they all have the same $3-$way average? – lulu Jul 18 '19 at 15:54
  • @Dr.SonnhardGraubner yeah exactly – roglol Jul 18 '19 at 15:55

2 Answers2

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You cannot be sure of what the average is. You only have $2$-equations with $3$-unknowns which means that there an infinite amount of $x,y,z$ that satisfy your initial equations. All of these triples are not guaranteed to have the same average. For example, the following triples satisfy your equations

$$ (400,400,1200), (300,500,1100) $$

however their averages are $\frac{2000}{3}, \frac{1900}{3}$ respectively

wjmccann
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If we add two equations, we can write that $$\frac{x+y+z}{3}=800-\frac{y}{3}$$ so the average depends on the value of $y$.

Vasili
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  • Technically since this is a system of $2$ equations with $3$ unknowns you can write the average as an amount dependent on any of the values – wjmccann Jul 22 '19 at 20:04