= Solution
With zero curvature and no <cosmological constant>, the present <normalization> requires $\Omega_M=1$. Thus $\dot a=H_0a^{-1/2}$, and choosing the big bang as $t=0$ gives
$$
\boxed{a(t)=\left(\frac{3H_0t}{2}\right)^{2/3},\qquad t_0=\frac2{3H_0}.}
$$
For an empty expanding model, $\Omega_k=1$ and the curvature must be negative. The equation becomes $\dot a=H_0$, giving
$$
\boxed{a(t)=H_0t,\qquad t_0=H_0^{-1}.}
$$
The latter is the Milne slicing of flat spacetime; its zero-scale-factor origin is not a matter-density curvature singularity. In the matter model, <gravity> decelerates expansion, so the earlier slope was greater than the present slope and the same present size was reached in less <time>. With the stipulated $H_0$, the two model ages are approximately $8.69$ and $13.0$ Gyr.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018-expansion-age-comparison.png]
{title=Equal present expansion rates give different ages when matter decelerates expansion}
{description=Both normalized scale factors reach one at the present and have the same slope there. The matter-only curve begins two thirds of a Hubble time before the present; the empty coasting curve begins one Hubble time before it.}
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