Atriangle has side lengths of 7, 13, and x what is the range of values for x?

Difference = 13-7 =6 
sum = 13+7 = 20
Looking at the attached graphic, when 2 sides are known,
the third side has to be greater than the difference and less than the sum
6 < third side < 20

Hint:  add 7 and 13.  You might want to mark off 7 units on a straight line, and then 13 more units on the same line.  Adding these two lengths together gives you the length of the third side (x) of a squashed (flat) triangle: 7+13=20.

Sketch this situation!

Now, leave the 13-unit-long side where it is and pivot one end of the 7-unit-long side about one end of the 13-unit-long side.  Will the 3rd side of your triangle now be greater or smaller than 20 units?  See whether you can figure out the shortest possible length that the third side could have.