Verified Answer :
see the picture attached to better understand the problem
Let
x----------> The distance from Chester to Durbin
we know that
\(10 \frac{3}{5} = \frac{(10*5+3)}{5} \\ \\ = \frac{53}{5} miles\)
\(12 \frac{1}{2} = \frac{(12*2+1)}{2} \\ \\ = \frac{25}{2} miles\)
\(10 \frac{3}{5} +12 \frac{1}{2} +x=35\)
subtitute
\(\frac{53}{5} + \frac{25}{2} +x=35\)
solve for x
\(x=35-( \frac{53}{5}+ \frac{25}{2} ) \\ \\ x=35- \frac{(2*53+5*25)}{10} \\ \\ x=35- \frac{231}{10} \\ \\ x=35-23.10 \\ \\ x=11.9\)
\(x=11 \frac{9}{10}\) miles
the answer is
\(11 \frac{9}{10}\) miles
Answer:
Durbin is \(11\frac{9}{10}\:miles\) far from Chester.
Step-by-step explanation:
Distance Between Alston and Barton = \(10\frac{3}{5}\:miles\)
Distance Between Barton and Chester = \(12\frac{1}{2}\:miles\)
Distance Between Alston and Durbin = \(35\:miles\)
Distance between Chester and Durbin = \(35-10\frac{3}{5}-12\frac{1}{2}\)
= \(35-\frac{53}{5}-\frac{25}{2}\)
= \(\frac{350-106-125}{10}\)
= \(\frac{119}{10}\)
= \(11\frac{9}{10}\:miles\)
Therefore, Durbin is \(11\frac{9}{10}\:miles\) far from Chester.