Since $M$ is the midpoint of $\overline{AB}$, the segment $\overline{AB}$ is twice the length of the segment $\overline{AM}.$ So, we have:

\begin{align*} 2AM &= AB \\[3pt] 2(3w+4) &= 7w+1 \\[3pt] 6w+8 &= 7w+1 \\[3pt] \end{align*}

Now, we subtract $6w$ from both sides of the equation:

\begin{align*} 6w+8-6w &= 7w+1-6w \\[3pt] 8 &= w+1 \\[3pt] \end{align*}

Next, we subtract $1$ from both sides:

\begin{align*} 8-1 &= w+1-1 \\[3pt] 7 &= w \\[3pt] w &= 7 \\[3pt] \end{align*}

Since $G$ is the midpoint of $\overline{EF}$, the segment $\overline{EF}$ is twice the length of the segment $\overline{EG}.$ So, we have:

\begin{align*} 2EG &= EF \\[3pt] 2(8-7y) &= 4-2y \\[3pt] 16-14y &= 4-2y \\[3pt] \end{align*}

Now, we add $14y$ to both sides of the equation:

\begin{align*} 16-14y+14y &= 4-2y+14y \\[3pt] 16 &= 4+12y \\[3pt] \end{align*}

Next, we subtract $4$ from both sides:

\begin{align*} 16-4 &= 4+12y-4 \\[3pt] 12 &= 12y \\[3pt] \end{align*}

Finally, we divide both sides by $12{:}$

\begin{align*} \dfrac{12}{12} &= \dfrac{12y}{12} \\[3pt] 1 &= y \\[5pt] y &= 1 \end{align*}

We have the following diagram.

Since $S$ is the midpoint of $\overline{PQ}$, the two segments $\overline{PS}$ and $\overline{SQ}$ have the same length. So, we have:

\begin{align*} PS &= SQ \\[5pt] 3x-1 &= 2x+3 \\[5pt] 3x-1-2x &= 2x+3-2x \\[5pt] x-1 &= 3 \\[5pt] x-1+1 &= 3+1 \\[5pt] x &= 4 \end{align*}

Therefore, $x = 4\,\textrm{in}.$

We have the following diagram.

Since $C$ is the midpoint of $\overline{AB}$, the segment $\overline{AB}$ is twice the length of the segment $\overline{AC}.$ So, we have:

\begin{align*} 2AC &= AB \\[5pt] 2(2x) &= 6x-1 \\[5pt] 4x &= 6x-1 \\[5pt] 4x-4x &= 6x-1-4x \\[5pt] 0 &= 2x-1 \\[5pt] 0+1 &= 2x-1+1 \\[5pt] 1 &= 2x \\[5pt] \dfrac{1}{2} &= \dfrac{2x}{2} \\[5pt] \dfrac{1}{2} &= x \\[5pt] x &= \dfrac 12 \end{align*}

Therefore, by plugging this value into the expression for $AB$, we obtain:

\begin{align*} AB &= 6x-1 \\[5pt] &= 6\cdot \left(\dfrac 12\right)-1 \\[5pt] &= 3-1 \\[5pt] &= 2 \end{align*}

The point $P$ corresponds to the number $x_1=-7$ while $Q$ corresponds to $x_2=3.$

To find the number $x_m$ corresponding to the midpoint $M$, we use the midpoint formula:

\begin{align*} x_m &= \dfrac{x_1+x_2}{2} \\[5pt] &= \dfrac{-7+3}{2} \\[5pt] &= \dfrac{-4}{2} \\[5pt] &= -2 \end{align*}

The point $A$ corresponds to the number $x_1=-2$ while $B$ corresponds to $x_2=4.$

To find the number $x_m$ corresponding to the midpoint $M$, we use the midpoint formula:

\begin{align*} x_m &= \dfrac{x_1+x_2}{2} \\[5pt] &= \dfrac{-2+4}{2} \\[5pt] &= \dfrac{2}{2} \\[5pt] &= 1 \end{align*}

The point $R$ corresponds to the number $x_1=-4$ while the midpoint $M$ corresponds to $x_m=-2.$

To find the number $x_2$ corresponding to the point $S$, we first set up an equation using the midpoint formula:

\begin{align*} x_m &= \dfrac{x_1+x_2}{2} \\[5pt] -2 &= \dfrac{-4+x_2}{2} \end{align*}

Now, we solve for $x_2\mathbin{:}$

\begin{align*} -2 &= \dfrac{-4+x_2}{2} \\[5pt] 2 \cdot (-2) &=-4 + x_2 \\[5pt] -4 &=-4 + x_2 \\[5pt] 0 &= x_2 \end{align*}

Therefore, the point $S$ corresponds to $x_2=0.$

The point $S$ corresponds to the number $x_2=-1$ while the midpoint $M$ corresponds to $x_m=-3.$

To find the number $x_1$ corresponding to the point $R$, we first set up an equation using the midpoint formula:

\begin{align*} x_m &= \dfrac{x_1+x_2}{2} \\[5pt] -3 &= \dfrac{x_1 + (-1)}{2} \end{align*}

Now, we solve for $x_1\mathbin{:}$

\begin{align*} -3 &= \dfrac{x_1 + (-1)}{2} \\[5pt] 2 \cdot (-3) &=x_1 -1 \\[5pt] -6 &=x_1 -1 \\[5pt] -6 +1 &= x_1 \\[5pt] -5 &= x_1 \end{align*}

Therefore, the point $R$ corresponds to $x_1=-5.$