Let l be the length of the rectangle and w be the width, since the length is five inches more than the width we can set the following equation:
[tex]l=w+5in\text{.}[/tex]Also, the perimeter of the rectangle is 50 inches, therefore:
[tex]2l+2w=50in\text{.}[/tex]Substituting the first equation in the second equation we get:
[tex]2(w+5in)+2w=50in\text{.}[/tex]Solving for w we get:
[tex]\begin{gathered} 2w+10in+2w=50in, \\ 4w=40in, \\ w=10in\text{.} \end{gathered}[/tex]Substituting w=10in in the first equation we get:
[tex]\begin{gathered} l=10in+5in, \\ l=15in\text{.} \end{gathered}[/tex]Answer: The length of the rectangle is 15 inches and the width is 10 inches.