Question

If x, y, and z represent three numbers such that y=z-4 and z=x-8, what is the result when the median of the numbers is subtracted from the average of the numbers?

Answer 1

\[\begin{cases}
x\\
y=z-4\\
z=x-8
\end{cases}~\Rightarrow~\color{red}{\begin{cases}x=x\\y=(x-8)-4\\z=x-8\end{cases}}\]
The average of x, y, and z is
\[average=\frac{x+y+z}{3}=\frac{x+(x-12)+(x-8)}{3}=\frac{3x-20}{3}\]
The median number is the "middle" number. So if you have \(x<y<z\), for example, then \(y\) is the median number.

In this case, judging by the red set of equations, \(z\) is the median number.
This is because
\[x-12<x-8<x~\iff~y<z<x\]
Now find the difference between the median and average.

Answer 2

Would you just add the 12 and the 8 on all sides?

Answer 3

thanks @SithsAndGiggles

Answer 4

No problem @Loser66.

@sakigirl, what do you mean by "all sides?"

The question wants you to find
\[average-median\]
which we've revealed to be equivalent to
\[\frac{3x-20}{3}-(x-8)\]

Answer 5

And so to subtract, would you multiply, 1/3 to x-8?

Answer 6

Not quite. You'd be missing a factor of 3 in the numerator. You have to multiply by 3/3:
\[\frac{3x-20}{3}-\frac{3(x-8)}{3}\]

Answer 7

So would it be -24/3 which is -8?

Answer 8

@SithsAndGiggles Sorry, you have no clue how incredibly dumb I feel, haha

Answer 9

Close! I think you made a small mistake with adding:
\[\frac{3x-20}{3}-\frac{3(x-8)}{3}\\
\frac{3x-20}{3}-\frac{3x-24}{3}\\
\frac{(3x-20)-(3x-24)}{3}\\
\frac{3x-20-3x+24}{3}\\
\frac{-20+24}{3}\\
~~~~~~~~~\vdots\]

Answer 10

@SithsAndGiggles Thank you :)

Answer 11

You're welcome!