<h3>
Answers:</h3>
<u> 14 </u> minutes below the average
<u> 6 </u> minutes above the average
=====================================================
Let A = average running time
Jacob's time was 4 minutes under the average, so his time is A-4 minutes. Whatever A is, subtract off 4, and you'll get Jacob's time.
Sophia and Jacob have a difference of 10 minutes. If J = Jacob's time and S = sophia's time, then S-J = 10 or J-S = 10 depending on who has the larger time.
We can use absolute value to ensure that whatever we pick (S-J or J-S) will be positive. So |S-J| = 10. Recall that absolute value represents distance on a number line. Negative distance isn't possible.
-------------
Let's plug in J = A-4 and solve for S
|S-J| = 10
|S - ( J )| = 10
|S - (A-4)|
|S-A+4| = 10
S-A+4 = 10 or S-A+4 = -10
S-A = 10-4 or S-A = -10-4
S-A = 6 or S-A = -14
S = A+6 or S = A-14
The equation S = A+6 shows Sophia is 6 minutes above the average
The equation S = A-14 shows Sophia is 14 minutes below the average
-------------
Let's pick some number for A that is over 14 minutes. Let's say the average running time is A = 20 minutes.
If A = 20, then Jacob's time is J = A-4 = 20-4 = 16
If the average running time is 20 minutes, then Jacob ran for 16 minutes.
If we subtract 10 from this, then J-10 = 16-10 = 6 is one possible time for Sophia. Notice how this is 14 minutes below the average (20-14 = 6)
If we add 10 to Jacob's time, then J+10 = 16+10 = 26, which is 6 minutes overage the average (20+6 = 26)
This is one numeric example, but you could use any value of A that you want as long as it's larger than 14. The reason A has to be larger than 14 is to ensure that Sophia's lower time value (A-14) is not negative. Having a time of zero is not feasible either.
