Math and Age.

Let the ages of Sam and Cindy be $S$ and $C$ , respectively, then in five years

$S+5=2(C+5)$ $\implies$ $S-2C=5$ $\rightarrow$ $\color{#D61F06}\boxed{1}$

and thirteen years ago

$S-13=3(C-13)$ $\implies$ $S-3C=-39+13$ $\implies$ $S-3C=-26$ $\rightarrow$ $\color{#D61F06}\boxed{2}$

Subtracting $\color{#D61F06}\boxed{2}$ from $\color{#D61F06}\boxed{1}$ , we get

$C=31$

It follows that $S$ is $5+(2)(31)=67$

$x$ years ago,

$S-x=4(C-x)$

$67-x=4(31-x)$

$67-x=124-4x$

$3x=57$

$\color{plum}\large{\boxed{x=19}}$

Let the ages of Sam and Cindy be $s$ and $c$ respectively. Then:

$\begin{cases} s+5 = 2(c+5) & \implies s = 2c + 5 \\ s-13 = 3(c-13) & \implies s = 3c -26 \end{cases}$

$\implies 2c+5 = 3c -26 \implies c = 31$ $\implies s = 2c+5 = 67$ .

Let $x$ years ago, Sam was four times as old as Cindy. Then

$\begin{aligned} s-x & = 4(c-x) \\ 67 - x & = 4(31-x) \\ 67 - x & = 124-4x \\ 3x & = 57 \\ \implies x & = \boxed{19} \end{aligned}$