Answer:
We'll first need to convert the mixed fractions to improper fractions and then subtract the distances.
1. Convert [tex]9 \frac{4}{5}[/tex] to an improper fraction:
[tex]9\dfrac{4}{5}= \dfrac{9x 5+4}{5}= \dfrac{45+4}{5}= \dfrac{49}{5}[/tex]
2. Convert [tex]6 \frac{3}{7}[/tex] to an improper fraction:
[tex]6 \frac{3}{7}= \frac{6x7+4}{7}=\frac{42+3}{7}= \frac{45}{7}[/tex]
Next, we need a common denominator to subtract these fractions. The least common multiple of 5 and 7 is 35.
3. Convert both fractions to have a denominator of 35:
[tex]\frac{49}{5}=\frac{49 x 7}{5x7}= \frac{343}{35}[/tex]
[tex]\frac{45}{7}=\frac{45x5}{7x5}=\frac{225}{35}[/tex]
4. Subtract the two fractions:
[tex]\frac{343}{35} - \frac{225}{35} = \frac{343 - 225}{35}= \frac{118}{35}[/tex]
Finally, convert the improper fraction back to a mixed number:
[tex]\frac{118}{35}= 3 \frac{13}{35}[/tex]
So, the difference in the distance needed to travel from City A to City B on the old road compared to the new road is [tex]3 \frac{13}{55}[/tex] miles.