The length of a rectangular picture frame is 5 inches (in.) less than twice its width. Its perimeter is 32 in. What are its dimensions?

Represent the length and width of the picture frame (given in inches). Let $x=\text{length of the picture frame}$ and $y=\text{width of the picture frame}\text{.}$

The length of the frame is 5 in. less than twice the width. Thus,

The perimeter of the frame is 32 in. Applying the formula for the perimeter of a rectangle, you will get

You now have a system of two linear equations in two variables.

To solve for y, substitute $2y-5$ for x in the second equation.

To solve for x, substitute 7 for y in $x=2y-5.$

Thus, the solution of the system is $(9,7).$

The length of the frame (which is 9 in.) is 5 in. less than twice the width, which is $2(7)=14\text{ in}\text{.}$ The perimeter of the frame is

The two numbers satisfy the conditions of the given problem. Thus, the length of the picture frame is 9 in. while its width is 7 in.