SOLUTION
Let the length of the rectangle be L, the width be w and the perimeter P
The length of a rectangle is two feet more than five times the width, this means
[tex]\begin{gathered} L=2+(5\times w) \\ L=2+5w \\ P=52 \end{gathered}[/tex]Also perimeter P of a rectangle is given as
[tex]P=2(L+w)[/tex]This becomes
[tex]\begin{gathered} P=2(L+w) \\ 52=2(L+w) \\ 52=2L+2w \\ \text{Now }L=2+5w \\ \text{substituting we have } \end{gathered}[/tex]That
[tex]\begin{gathered} 52=2L+2w \\ 52=2(2+5w)+2w \\ 52=4+10w+2w \\ \text{collecting like terms } \\ 52-4=10w+2w \\ 48=12w \\ w=\frac{48}{12} \\ w=4 \end{gathered}[/tex]Hence the width is 4
The length becomes
[tex]\begin{gathered} L=2+5w \\ L=2+5\times4 \\ L=2+20 \\ L=22 \end{gathered}[/tex]And the length is 22
Hence the dimensions of the rectangle is 22 by 4
