You commute 56 miles one way to work. The trip to work takes 10 minutes longer than the trip home. Your average speed on the trip home is 8 miles per hour faster. What is your average speed on the trip home?
I got to this point
xy=56
(x-8)(y+10)=56
I couldn't get any further than that. I'm not sure where to go with those equations.
Ok, then x=56/y
(56/y -8)*(y+10/60)=56
multiply both sides by y.
y(56/y -8)*(y+10/60)=56y
(56 -8y)*(y+10/60)=56y
FOIL that, and solve the quadratic for y. ONce you get that, you can solve for x.
Note:
The fraction 10/60 is important..the ten was in minutes, and it has to be converted to hours, as speed was given in km/hr (remember the units given for 8).
To solve this problem, let's first simplify the equations you have:
From the first equation, xy = 56.
From the second equation, (x-8)(y+10) = 56.
Now, let's work on solving these equations step by step:
1. Start with the first equation: xy = 56.
Since we know that x and y represent the distances for the trip to work and trip back home respectively, we can rewrite this equation as:
x * y = 56.
2. Next, let's focus on the second equation: (x-8)(y+10) = 56.
Expanding this equation, we get:
xy + 10x - 8y - 80 = 56.
Rearranging the terms, we have:
xy + 10x - 8y = 136.
3. Now, we can substitute the value of xy from the first equation into the second equation to eliminate one variable:
56 + 10x - 8y = 136.
4. Simplifying this equation further, we get:
10x - 8y = 80.
5. Rearrange this equation by isolating one variable. Let's solve for x:
10x = 8y + 80.
Divide both sides of the equation by 10:
x = (8y + 80) / 10.
6. We can use this expression for x and substitute it back into the first equation:
[(8y + 80) / 10] * y = 56.
Now, cross-multiply to get:
(8y + 80) * y = 560.
Expanding this equation gives us:
8y^2 + 80y = 560.
7. Simplify further:
Divide both sides of the equation by 8 to get:
y^2 + 10y = 70.
8. Rearrange this equation to solve for y. We have a quadratic equation, so let's set it equal to zero:
y^2 + 10y - 70 =0.
9. Factor this quadratic equation:
(y + 14)(y - 4) = 0.
This gives us two possible values for y:
y = -14 or y = 4.
10. Since distance cannot be negative, we discard y = -14, and the valid solution is y = 4.
Now, substitute this value back into the x expression we obtained earlier:
x = (8y + 80) / 10.
x = (8*4 + 80) / 10.
x = 112 / 10.
x = 11.2.
11. We now know that x = 11.2 and y = 4, representing the distance for the trip to work and trip back home, respectively.
To find the average speed on the trip home, we need to calculate the time taken for the trip back home.
The time taken for the trip to work (x) can be calculated by dividing the distance by the speed:
Time taken for the trip to work = distance / speed = 11.2 / y.
The time taken for the trip back home (y) is 10 minutes (1/6 hour) less:
Time taken for the trip back home = distance / (speed + 8) = 4 / (y + 8).
12. Now, let's use these expressions to find the average speed on the trip back home.
Average speed = distance / time.
Average speed on the trip back home = 4 / [(4 / (y + 8)) - (1/6)].
13. Substitute the value of y into this expression:
Average speed on the trip back home = 4 / [(4 / (4 + 8)) - (1/6)].
Average speed on the trip back home = 4 / [(4 / 12) - (1/6)].
Average speed on the trip back home = 4 / (1/3 - 1/6).
Average speed on the trip back home = 4 / (2/6 - 1/6).
Average speed on the trip back home = 4 / (1/6).
Average speed on the trip back home = 24 mph.
Therefore, the average speed on the trip back home is 24 miles per hour.