To solve this question we will use the following diagram:
The area of the big rectangle minus the area of the small rectangle is:
[tex]14in\times16in-48in^2=176in^2.[/tex]Since the frame has a uniform width, then:
[tex]16x+16x+(14-2x)x+(14-2x)x=176.[/tex]Solving the above equation for x we get:
[tex]\begin{gathered} 32x+28x-4x^2=176, \\ 15x-x^2-44=0. \\ x^2-15x+44=0, \\ (x-4)(x-11)=0. \end{gathered}[/tex]Then:
[tex]x=4\text{ or x=11,}[/tex]but 11 is not a solution with real meaning, therefore x=4.
Answer: The width is 4 in.
