Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman $\alpha$ and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from $z \equiv \lambda_\mathrm{obs}/\lambda_\mathrm{rest} - 1$),
$$
\begin{array}{rcl}
z_{\mathrm{Ly}\alpha} & = & \frac{317.7\,\mathrm{nm}}{125.67\,\mathrm{nm}} & \simeq & 1.6134 \\
z_\mathrm{C\,IV} & = & \frac{404.7\,\mathrm{nm}}{154.9\,\mathrm{nm}} & \simeq & 1.6127.
\end{array}
$$
That is, Lyman $\alpha$ is slightly more redshifted than C IV. This is often seen, and is due to Lyman $\alpha$ scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by $z\simeq6$, the blue part of the spectrum is completely gone (this is the so-called Gunn-Peterson trough).
Relating the widths $\Delta\lambda$ of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by $v/c = \Delta\lambda / \lambda$, so the velocities are
$$
\begin{array}{rcl}
v_{\mathrm{Ly}\alpha} & = & c \frac{10\,\mathrm{nm}}{317.7\,\mathrm{nm}} & \simeq & 9\,400\,\mathrm{km}\,\mathrm{s}^{-1} \\
v_\mathrm{C\,IV} & = & c \frac{20\,\mathrm{nm}}{404.7\,\mathrm{nm}} & \simeq & 14\,800\,\mathrm{km}\,\mathrm{s}^{-1}.
\end{array}
$$
From simple dynamics, this can be converted to a distance $r$ from the black hole, knowing its mass $M_\bullet$:
$$
\begin{array}{rcl}
r & = & \frac{G M_\bullet}{v^2} & \Rightarrow \\
r_{\mathrm{Ly}\alpha} & \simeq & 0.05\,\mathrm{pc} & \simeq & 58\,\mathrm{light\text{-}days} \\
r_{\mathrm{C\,IV}} & \simeq & 0.02\,\mathrm{pc} & \simeq & 23\,\mathrm{light\text{-}days},
\end{array}
$$
which is consistent with the fact that these lines are formed in the so-called broad line region which has dimension from tens of light-days to several parsec (in contrast, the narrow line region has dimensions of several 100s to ~1000 light-years).
