a two part canoe trip,the first part of the trip the canoist traveled 58 miles.Second part of the trip he traveled 18 miles at 5mph slower.the total time for trip was 4 hours.What was the speed of both parts of the trip?

faster speed --- x mph

slower speed ---- x-5 mph

time at slower speed = 58/x
time at faster speed = 18/(x-5)
58/x + 18/(x-5) = 4
times x(x-5)

58(x-5) + 18x = 4x(x-5)
58x - 290 + 18x = 4x^2 - 20x
4x^2 -96x + 290 = 0
2x^2 - 48x + 145 = 0

x = (48 ± √1144)/4
= appr. 20.45 mph or 3.544 mph

if x = 20.45 , then x-5 = 15.45 mph
if x = 3.544 mph , then x-5 is a negative, so NO GOOD

Ridiculous answer!
I have done my share of canoeing, there is no way you can paddle a canoe at 20 mph!!!!
Absurd to canoe 76 miles in 4 hours.

Thanks

To find the speed of both parts of the trip, we can use the formula:

Speed = Distance / Time.

Let's assume the speed for the first part of the trip is S mph.

For the first part of the trip, the distance covered is 58 miles. Therefore, the time taken can be calculated as:

Time = Distance / Speed
Time = 58 miles / S mph
Time = 58/S hours.

Now, for the second part of the trip, the canoeist travels at a speed 5 mph slower than the first part. So, the speed for the second part of the trip is (S - 5) mph.

For the second part, the distance covered is 18 miles. Therefore, the time taken can be calculated as:

Time = Distance / Speed
Time = 18 miles / (S - 5) mph
Time = 18/(S - 5) hours.

The total time for the trip is given as 4 hours. Since the first and second parts of the trip are done consecutively, we can add the time taken for each part:

Time for first part + Time for second part = Total time
58/S + 18/(S - 5) = 4.

Now, we can solve this equation to find the value of S, which is the speed for the first part of the trip:

58/S + 18/(S - 5) = 4.

To simplify this equation, we can clear the fractions by multiplying both sides by the least common denominator (LCD), which is S(S - 5):

(S(S - 5))(58/S + 18/(S - 5)) = 4(S(S - 5)).

This simplifies to:

58(S - 5) + 18S = 4S(S - 5).

Now, we can further simplify:

58S - 290 + 18S = 4S^2 - 20S.

Rearranging and simplifying:

4S^2 - 20S - 76S + 290 + 0 = 0,
4S^2 - 96S + 290 = 0.

Next, we can solve this quadratic equation using either factoring, completing the square, or the quadratic formula:

Solving for S using the quadratic formula, where a = 4, b = -96, and c = 290:

S = (-b ± √(b^2 - 4ac)) / 2a.

S = (-(-96) ± √((-96)^2 - 4(4)(290))) / (2(4)).
S = (96 ± √(9216 - 4640)) / 8.
S = (96 ± √4576) / 8.
S = (96 ± 67.67) / 8.

Now, we can find the possible values for S:

S1 = (96 + 67.67) / 8 ≈ 16.96 mph,
S2 = (96 - 67.67) / 8 ≈ 3.16 mph.

Therefore, the two possible speeds for the first part of the trip are approximately 16.96 mph and 3.16 mph.