Respuesta :

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.

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