Dave drove to his friends house and back. it took one hour longer to go there than it did to come back. the average speed on the trip there was 25mph. the average speed on the way back was 30mph. how many hours did the trip there take? (urgent)

Remark

You know the most about the trip on the way back and the total of the two trips.

Way Back

r = 30 mph

d = distance traveled

t = t - 1 where t = the time to go there which we know nothing about

Total Distance

2d = 25 mph * (t + t - 1)

2d = 25 * (2t - 1)

2d = 50t  - 25                 Divide by 2

2d = 50t/2 - 25/2

d = 25t - 12.5

Equation coming back

d = 30*(t - 1)

d = 30t - 30

Comment

Since the distances are the same, equate them.

Solution

25t - 12.5 = 30t - 30                     Subtract 25t from both sides

25t - 25t -12.5 = 30t - 25t - 30   Combine like terms

-12.5 = 5t - 30                              Add 30 to both sides

-12.5 + 30 = 5t - 30 + 30             Combine

17.5 = 5t                                       Divide by 5

17.5/5 = t                                      Divide and Switch

t = 3.5

t is the time going there

t - 1 is the time coming back

Total time = 2*t - 1 = 2*3.5 - 1 = 6

The total time is 6 hours.   Answer

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?