Context: Line $k$ contains the points $(-3, -48)$, $(v, 0)$, and $(5, 56)$. Since $(v, 0)$ is where the line crosses the $x$-axis, we need that $x$-value.

Answer: 9/13

Why: Use the two fully known points to find the slope: $\dfrac{56 - (-48)}{5 - (-3)} = \dfrac{104}{8} = 13$. Point-slope form through $(5, 56)$: $y = 13(x - 5) + 56 = 13x - 65 + 56 = 13x - 9$. The point $(v, 0)$ has $y = 0$, so $13v - 9 = 0$, which gives $v = \dfrac{9}{13}$.

Verification: The line $y = 13x - 9$ also fits the remaining point: $13(-3) - 9 = -39 - 9 = -48$. And $13\left(\dfrac{9}{13}\right) - 9 = 9 - 9 = 0$, as required.

Common slips: Computing the slope upside down as $\dfrac{8}{104}$, or making a sign error in the line equation and solving $13v + 9 = 0$, which gives $-\dfrac{9}{13}$ instead.