Jake's morning run was a distance of 9.3 miles. He ran back and forth on one long straight street. He started at his house and ran to the school 4 miles away. Then he ran to the library, and then he ran back to his house. What are possible distances the library could be from the house?

the distances can be 0.65 miles (in the opposite direction to the school) or 4.65 miles (in the same direction as the school)

Step-by-step explanation:

Ok, the data that we have is:

Total distance = 9.3 mi.

The travel is:

House to school = 4 mi.

school to library = A

library to house = B

Now, we have that:

4mi + A + B = 9.3mi.

We have three possibilities:

1) The order of locations is: house, library, school

The travel from: school to library + library to house is equivalent to a travel between the school to the house = 4mi.

Then we have A + B = 4mi

4mi + A + B = 8mi ≠ 9.3mi

Then the library can not be between the house and the school.

2) The order of locations is: house, school, library.

In this case we have that the distance between the library and the house is equal to the distance between the house and the school plus the distance between the school and the library, then:

4mi + A = B.

We can replace this in our original equation:

4mi + A + B = 9.3mi

4mi  + A + (4mi + A) = 9.3mi

8mi + 2*A = 9.3mi

2*A = 9.3mi - 8mi = 1.3mi

A = 1.3mi/2 = 0.65mi

Then the distance between the house and the library is:

The 4 miles between the house and the school, plus the 0.65 miles between the school and the library:

Distance = 4mi + 0.65mi. = 4,65mi

3) The third case is when the order of the locations is:

Library, house, school.

Then the distance between the house and the library is equal to the distance between the school and the library minus the distance between the house and the school, this is:

A - 4mi = B

Now we can replace this in our distance equation:

4mi + A + B = 9.3mi

4mi + A + (A - 4mi) = 9.3 mi

2A = 9.3mi

A = 9.3mi/2 = 4.65mi

Then the distance between the house and the library is:

B = A - 4mi = 4.65mi - A = 0.65mi

Then the distance between the house and the library is 0.65 miles in this case.

chad should not have included the values (1, 2) and (3, 6) because that would mean he only had half a healing card.