Answer: The possible distances are 0.65 miles (away from the school) or 4.65 miles (towards the school).
Step-by-step explanation:
The information we have is as follows:
Total distance = 9.3 miles.
Journey:
House to school = 4 miles.
School to library = A
Library to house = B
From this, we know that:
4 miles + A + B = 9.3 miles.
We have three scenarios:
1) If the sequence is: house, library, school
The distance from school to library plus library to house would equal the distance from school to house = 4 miles.
Thus, A + B = 4 miles
4 miles + A + B = 8 miles ≠ 9.3 miles
Therefore, the library cannot be located between the house and school.
2) If the order is: house, school, library.
In this case, the distance from library to house would be the sum of the distance from house to school and from school to library:
4 miles + A = B.
Substituting this into our original equation:
4 miles + A + B = 9.3 miles
4 miles + A + (4 miles + A) = 9.3 miles
8 miles + 2*A = 9.3 miles
2*A = 9.3 miles - 8 miles = 1.3 miles
A = 1.3 miles / 2 = 0.65 miles
Thus, the distance from house to library is:
The 4 miles to school, plus the 0.65 miles from school to library:
Distance = 4 miles + 0.65 miles = 4.65 miles.
3) In the last case, the sequence is:
Library, house, school.
Here, the distance from house to library equals the difference between the distance from school to library and the distance from house to school:
A - 4 miles = B
Let's substitute this into our distance equation:
4 miles + A + B = 9.3 miles
4 miles + A + (A - 4 miles) = 9.3 miles
2A = 9.3 miles
A = 9.3 miles / 2 = 4.65 miles
Then the distance from house to library is:
B = A - 4 miles = 4.65 miles - A = 0.65 miles
So, the distance from house to library is 0.65 miles in this scenario.
