Answer:
![k=7; k=13](https://tex.z-dn.net/?f=k%3D7%3B%20k%3D13)
Fast and loose:
If you stare at it for a while, you can see that the LHS is the distance between k and 10. Basically, you've asked to pick k so you get two numbers that are 3 apart from 10. And these two numbers are 7 to the left, and 13 to the right.
Rigorously
Split it in 2 situations, depending if the quantity inside the absolute value is greater or smaller than zero.
If it's greater or equal than zero:
![\left \{ {{k-10\geq 0 } \atop {k-10 = 3} \right. \\\left \{ {{k\geq 10 } \atop {k = 13} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bk-10%5Cgeq%200%20%7D%20%5Catop%20%7Bk-10%20%3D%203%7D%20%5Cright.%20%5C%5C%5Cleft%20%5C%7B%20%7B%7Bk%5Cgeq%2010%20%7D%20%5Catop%20%7Bk%20%3D%2013%7D%20%5Cright.)
And since 13 is indeed greater than 10, it's a valid solution, so you get the first one. For the second, you assume negative values for the quantity inside, and change sign:
![\left \{ {{k-10 < 0 } \atop {k-10 = -3} \right. \\\left \{ {{k < 10 } \atop {k= 7} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bk-10%20%3C%200%20%7D%20%5Catop%20%7Bk-10%20%3D%20-3%7D%20%5Cright.%20%5C%5C%5Cleft%20%5C%7B%20%7B%7Bk%20%3C%2010%20%7D%20%5Catop%20%7Bk%3D%207%7D%20%5Cright.)
In this case, 7 is a valid solution too since it's less than 10