The perimeter of a fence in the shape of a square needs to be less than 30 feet in order to have enough fencing to go around. In interval notation, how much fence could be used for the lengths of each side?

The length would be 22.

Step-by-step explanation (written by my sis): The first step in a problem like this (after sketching the problem out for yourself, which very often helps) is to assign variables.  Let's call the length x, and the width y.

There are two sections that are y long, and one that's x long.  Adding up the total length of the fence, then, gives us x + y + y, or x + 2y, which should be equal to 44, since that's how much fence we have.  Our first equation is thus

x + 2y = 44

Our second equation, since the length is supposed to be twice the width, is x = 2y.

This is enough to solve.  We can substitute this equation for x into the first equation to get

2y + 2y = 44

4y = 44

y = 11

And then x = 2(11) = 22

So the length of the fence is 22, and the width (though they don't actually ask for it) is 11.

Check the  22 + 11 + 11 is 44, as we want, and 22 is twice as great as 11, as we want.  So the answer is valid.

associative property of multiplication