uations

28 x + 10 = 19 y + 2 = 15 z + 4.

To solve the equations 28 x + 10 = 19 y + 2, or y = x + (9 x + 8) / 19, let
m = (9 x + 8) / 19, we have then x = 2 m + (m - 8) / 9.

Let (m - 8) / 9 = m'; then m = 9 m' + 8; hence

x = 18 m' + 16 + m' = 19 m' + 16 . . . (1).

Again, since 28 x + 10 = 15 z + 4, we have

15 z = 28 x + 6, or z = 2 x - (2 x - 6) / 15.

Let (2 x - 6) / 15 = n; then 2 x = 15 n + 6, and x = 7 n + 3 + n / 2.

Let n / 2 = n'; then n = 2 n'; consequently

x = 14 n' + 3 + n' = 15 n' + 3 . . . (2).

Equating the above two values of x, we have

15 n' + 3 = 19 m' + 16; whence n' = m' + (4 m' + 13) / 15.

Let (4 m' + 13) / 15 = p; we have then

4 m' = 15 p - 13, and m' = 4 p - (p + 13) / 4.

Let (p + 13) / 4 = p'; then p = 4 p' - 13;

whence m' = 16 p' - 52 - p' = 15 p' - 52.

Now in this equation p' may be any number whatever, provided 15 p' exceed
52. The smallest value of p' (which is the one here wanted) is therefore 4;
for 15 x 4 = 60. Assuming therefore p' = 4, we have m' = 60 - 52 = 8; and
consequently, since x = 19 m' + 16, x = 19 x 8 + 16 = 168. The number
required is consequently 28 x 168 + 10 = 4714.

Having found the number 4714 for the first of the era, the correspondence
of the years of the era and of the period is as follows:--

Era, 1, 2, 3, ... x,
Period, 4714, 4715, 4716, ... 4713 + x;

from which it is evident, that if we take P to represent the year of the
Julian period, and x the corresponding year of the Christian era, we shall
have

P = 4713 + x, and x = P - 4713.

With regard to the numeration of the years previous to the commencement of
the era, the practice is not uniform. Chronologists, in general, reckon the
year preceding the first of the era -1, the next preceding -2, and so on.
In this case

Era, -1, -2, -3, ... -x,
Period, 4713, 4712, 4711, ... 4714 - x;

whence

P = 4714 - x, and x = 4714 - P.

But astronomers, in order to preserve the uniformity of computation, make
the series of years proceed without interruption, and reckon the year
preceding the first of the era 0. Thus

Era, 0, -1, -2, ... -x,
Period, 4713, 4712, 4711, ... 4713 - x;

therefore, in this case

P = 4713 - x, and x = 4713 - P.

_Reformation of the Calendar._--The ancient church calendar was founded on
two suppositions, both erroneous, namely, that the year contains 3651/4 days,