[Work Included] In ABC, centroid D is on median AM. AD = x + 5 and DM = 2x – 1 Find AM.

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[Work Included] In ABC, centroid D is on median AM. AD = x + 5 and DM = 2x – 1 Find AM.

Mathematics
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It looks more like an orthocenter, but for the sake of it, let's say it is a centroid.
\[ x + 5 = 2x - 1\]\[5 + x - 2x = 2x - 2x - 1\]\[5 - x = -1\]

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Subtract five from both sides. \[-x = -1 - 5\]\[-x = -6\]
\[x = 6\]
Now I can plug in the missing numbers. \[6 + 5 = 2(6) - 1\]\[11 = 12 - 1\]\[= 11\]
I think you're incorrect in assuming that \(x+5\) must equal \(2x-1\). The centroid of a triangle is located at a point 2/3 of the way along the median, and not 1/2 of the way along the median.
In that case, I'll tweak a few things here. \[x + 5 = 2(2x - 1)\]\[x + 5 = 4x - 2\] Subtracting 4 from both sides.
\[5 - 3x = -2\]\[-3x = -2 - 5\]\[-3x = -7\]
\[x = \frac{ -7 }{ -3 }\]
That looks good to me. You can simplify that to \(\dfrac{7}{3}\) if you want.
You're welcome.
@KingGeorge The answer was really 11. Guess I should have trusted my gut, ha.
Oops :( I guess that would be my fault.
However, I think we just forgot the last step in the problem. We found \(x=7/3\), but AM is equal to \((x+5)+(2x-1)\). If we plug in our value for \(x\) into this, we get\[\frac{7}{3}+4+\frac{14}{3}=\frac{21}{3}+4=7+4=11\]

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