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There is this question I was approached with and I am absolutely stumped on where to even begin to solve it. It goes like this:

Based off the following 4 equations, find the average of $x_1$, $x_2$, $x_3$, and $x_4$.

$$x_1 + \frac{x_2+x_3+x_4}{3} = 25$$

$$x_2 + \frac{x_1+x_3+x_4}{3} = 37$$

$$x_3 + \frac{x_1+x_2+x_4}{3} = 43$$

$$x_4 + \frac{x_1+x_2+x_3}{3} = 51$$

However I'm unsure of how to move on from here or if I am even going in the right direction.

Backslash
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  • For more infor than just the average (see @Twieceler's answer below), you could just solve the system of equations using Gaussian Elimination. – gt6989b Feb 20 '13 at 21:51
  • You could multiply each equation by 3 on both sides to remove the fractions and then row reduce it to get somewhere. – JB King Feb 20 '13 at 21:55
  • But Gaussian elimination is a lot more work than needed to get the requested answer. – Ross Millikan Feb 20 '13 at 23:39

1 Answers1

4

Hint:

Sum the left-hand sides and the right-hand sides of the 4 equations. What do you get?

Twiceler
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  • After doing so, I ended up with $6x_1 + 6x_2 + 6x_3 + 6x_4 = 156$, and from there I divided everything by $6$, and then divided everything by $4$. Was this the right approach? The final answer was $6\frac{1}{2}$. – Backslash Feb 20 '13 at 23:33
  • @RossMillikan How would I get $5\frac{1}{3}$ as the coefficient?I found out that $6$ is definitely not right however after recalculating I am getting a coefficient of $4$. – Backslash Feb 20 '13 at 23:56
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    @Qasim: I didn't read carefully enough. You get a full $x_1$ from the first one and three thirds from the last three, so a total of $2$. Similarly for the others. – Ross Millikan Feb 20 '13 at 23:58