Answer :
To find the difference in the time it takes Jesse to travel upstream and downstream, we start with the given expression:
[tex]\[ \frac{8}{5-c} - \frac{8}{5+c} \][/tex]
We need to simplify this expression.
Step 1: Identify a common denominator
The common denominator for the two fractions is \((5 - c)(5 + c)\).
Step 2: Rewrite each fraction with the common denominator
To achieve this, we multiply the numerator and the denominator of each fraction by the necessary term to obtain the common denominator:
[tex]\[ \frac{8}{5-c} \cdot \frac{5+c}{5+c} = \frac{8(5+c)}{(5-c)(5+c)} \][/tex]
[tex]\[ \frac{8}{5+c} \cdot \frac{5-c}{5-c} = \frac{8(5-c)}{(5-c)(5+c)} \][/tex]
Step 3: Combine the fractions over the common denominator
Now we can subtract the two fractions:
[tex]\[ \frac{8(5+c)}{(5-c)(5+c)} - \frac{8(5-c)}{(5-c)(5+c)} \][/tex]
Since they have the same denominator, we can combine the numerators:
[tex]\[ \frac{8(5+c) - 8(5-c)}{(5-c)(5+c)} \][/tex]
Step 4: Simplify the numerator
Distribute the 8 in each part of the numerator:
[tex]\[ 8(5+c) = 40 + 8c \][/tex]
[tex]\[ 8(5-c) = 40 - 8c \][/tex]
Substitute these back into the expression:
[tex]\[ \frac{40 + 8c - (40 - 8c)}{(5-c)(5+c)} = \frac{40 + 8c - 40 + 8c}{(5-c)(5+c)} \][/tex]
Combine like terms in the numerator:
[tex]\[ \frac{40 + 8c - 40 + 8c}{(5-c)(5+c)} = \frac{16c}{(5-c)(5+c)} \][/tex]
Step 5: Simplify the denominator
Recognize that \((5-c)(5+c)\) is a difference of squares:
[tex]\[ (5-c)(5+c) = 25 - c^2 \][/tex]
Substitute this back into the expression:
[tex]\[ \frac{16c}{25 - c^2} \][/tex]
Thus, the simplified form of the expression representing the time difference is:
[tex]\[ - \frac{16c}{c^2 - 25} \][/tex]
Therefore, the difference in the time it takes Jesse to travel upstream and downstream is:
[tex]\[ - \frac{16c}{c^2 - 25} \][/tex]
[tex]\[ \frac{8}{5-c} - \frac{8}{5+c} \][/tex]
We need to simplify this expression.
Step 1: Identify a common denominator
The common denominator for the two fractions is \((5 - c)(5 + c)\).
Step 2: Rewrite each fraction with the common denominator
To achieve this, we multiply the numerator and the denominator of each fraction by the necessary term to obtain the common denominator:
[tex]\[ \frac{8}{5-c} \cdot \frac{5+c}{5+c} = \frac{8(5+c)}{(5-c)(5+c)} \][/tex]
[tex]\[ \frac{8}{5+c} \cdot \frac{5-c}{5-c} = \frac{8(5-c)}{(5-c)(5+c)} \][/tex]
Step 3: Combine the fractions over the common denominator
Now we can subtract the two fractions:
[tex]\[ \frac{8(5+c)}{(5-c)(5+c)} - \frac{8(5-c)}{(5-c)(5+c)} \][/tex]
Since they have the same denominator, we can combine the numerators:
[tex]\[ \frac{8(5+c) - 8(5-c)}{(5-c)(5+c)} \][/tex]
Step 4: Simplify the numerator
Distribute the 8 in each part of the numerator:
[tex]\[ 8(5+c) = 40 + 8c \][/tex]
[tex]\[ 8(5-c) = 40 - 8c \][/tex]
Substitute these back into the expression:
[tex]\[ \frac{40 + 8c - (40 - 8c)}{(5-c)(5+c)} = \frac{40 + 8c - 40 + 8c}{(5-c)(5+c)} \][/tex]
Combine like terms in the numerator:
[tex]\[ \frac{40 + 8c - 40 + 8c}{(5-c)(5+c)} = \frac{16c}{(5-c)(5+c)} \][/tex]
Step 5: Simplify the denominator
Recognize that \((5-c)(5+c)\) is a difference of squares:
[tex]\[ (5-c)(5+c) = 25 - c^2 \][/tex]
Substitute this back into the expression:
[tex]\[ \frac{16c}{25 - c^2} \][/tex]
Thus, the simplified form of the expression representing the time difference is:
[tex]\[ - \frac{16c}{c^2 - 25} \][/tex]
Therefore, the difference in the time it takes Jesse to travel upstream and downstream is:
[tex]\[ - \frac{16c}{c^2 - 25} \][/tex]
