Since this can't happen, let's suppose that P=150, then:
[tex]\begin{gathered} P=2l+2w \\ \Rightarrow150=2(50)+2w \\ \Rightarrow150=100+2w \end{gathered}[/tex]Now we have to move the 100 to the other side of the equation with a negative sign to get this:
[tex]\begin{gathered} 150=100+2w \\ \Rightarrow150-100=2w \\ \Rightarrow50=2w \end{gathered}[/tex]Finally, to get w, we move the 2 that's multiplying to the other side dividing the 50:
[tex]\begin{gathered} 50=2w \\ \Rightarrow\frac{50}{2}=w \\ \Rightarrow w=25 \end{gathered}[/tex]Therefore, the width of the rectangle would be 25 ft if the perimeter is 150ft, and we can see how the rectangle would look: