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