jmc

Let $l$ represent the length of the rectangle and $w$ represent the width so that $l=3x$ and $w=x+5$. Since the area of the rectangle equals its perimeter, we have that $l\times w=2l+2w$. We can then substitute $3x$ back in for $l$ and $x+5$ in for $w$ to get $$\begin{array}{rl}& (3x)(x+5)=2(3x)+2(x+5)\\ \Rightarrow & 3x^2+15x=6x+2x+10\\ \Rightarrow & 3x^2+7x-10=0\\ \Rightarrow & (x-1)(3x+10)=0.\end{array}$$Solving this equation, we get that the two possible values of $x$ are $x=1$ and $x=-\frac{10}{3}$. However, both the length $3x$ and the width $x+5$ must be positive, so the only solution is $x=\boxed{1}$.