How far is it to his office?

rpenner · Oct 10, 2014

Motor Daddy said: ↑

rpenner, How did you figure out that he walked for 1km on the original trip of 25 minutes?
Click to expand...

Well, it turns out I could have relied on the OP for that, but I also calculated it myself in post #9#10.

rpenner said: ↑

Define:
x = distance to office, in km (What we want to solve for)
y = normal distance walked, in km (what we don't care about)
...
\(\begin{pmatrix} \frac{1}{40} & \frac{1}{6} - \frac{1}{40} \\ \frac{1}{40} & \frac{3}{6} - \frac{3}{40} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{25}{60} \\ \frac{25}{60} + \frac{17}{60} \end{pmatrix} \)
Since \(\begin{pmatrix} \frac{1}{40} & \frac{1}{6} - \frac{1}{40} \\ \frac{1}{40} & \frac{3}{6} - \frac{3}{40} \end{pmatrix} = \begin{pmatrix} \frac{1}{40} & \frac{17}{120} \\ \frac{1}{40} & \frac{17}{40} \end{pmatrix} \) and \( \begin{pmatrix} 60 & -20 \\ -\frac{60}{17} & \frac{60}{17} \end{pmatrix} \begin{pmatrix} \frac{1}{40} & \frac{17}{120} \\ \frac{1}{40} & \frac{17}{40} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) we have
\( \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 60 & -20 \\ -\frac{60}{17} & \frac{60}{17} \end{pmatrix} \begin{pmatrix} \frac{25}{60} \\ \frac{25}{60} + \frac{17}{60} \end{pmatrix} = \begin{pmatrix} 25 - 14 \\ -\frac{25}{17} + \frac{42}{17} \end{pmatrix} = \begin{pmatrix} 11 \\ 1 \end{pmatrix} \)
Click to expand...

x = distance to office, in km = 11 km
y = normal distance walked, in km = 1

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Motor Daddy · Oct 10, 2014

Thanks. As it turns out origin posted in post #9, not you.

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rpenner · Oct 10, 2014

Oh, right. Post numbers are at the bottom now.

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How far is it to his office?

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How far is it to his office?

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