We have a rectangle where the length L is 4 inches longer than twice the width W.
Twice the width is 2W, so we can write the length L as:
[tex]L=2W+4[/tex]The perimeter is 50 inches.
It is also two times the sum of the length L and the width W, so we can write:
[tex]2(L+W)=50[/tex]We can replace the length L in the perimeter equation and then solve for W as:
[tex]\begin{gathered} 2(L+W)=50 \\ 2\lbrack(2W+4)+W\rbrack=50 \\ 2W+4+W=\frac{50}{2} \\ 3W+4=25 \\ 3W=25-4 \\ 3W=21 \\ W=\frac{21}{3} \\ W=7 \end{gathered}[/tex]Knowing the width, we can calculate the length L as:
[tex]\begin{gathered} L=2W+4 \\ L=2(7)+4 \\ L=14+4 \\ L-18 \end{gathered}[/tex]Answer: the rectangle is 18 inches long and 7 inches wide.
