imc

If the two equal values are $3$ and $x+2$, then $x=1$. Also, $y-4\le 3$ because $3$ is the common value. Solving for $y$, we get $y\le 7$. Therefore the portion of the line $x=1$ where $y\le 7$ is part of $S$. This is a ray with an endpoint of $(1,7)$. Similar to the process above, we assume that the two equal values are $3$ and $y-4$. Solving the equation $3=y-4$ then $y=7$. Also, $x+2\le 3$ because 3 is the common value. Solving for $x$, we get $x\le 1$. Therefore the portion of the line $y=7$ where $x\le 1$ is also part of $S$. This is another ray with the same endpoint as the above ray: $(1,7)$. If $x+2$ and $y-4$ are the two equal values, then $x+2=y-4$. Solving the equation for $y$, we get $y=x+6$. Also $3\le y-4$ because $y-4$ is one way to express the common value. Solving for $y$, we get $y\ge 7$. We also know $3\le x+2$, so $x\ge 1$.Therefore the portion of the line $y=x+6$ where $y\ge 7$ is part of $S$ like the other two rays. The lowest possible value that can be achieved is also $(1,7)$. Since $S$ is made up of three rays with common endpoint $(1,7)$, the answer is $\boxed{\text{three rays with a common endpoint}}$