Page:Scientific Memoirs, Vol. 3 (1843).djvu/689

as an example the resolution of two equations of the first degree with two unknown quantities. Let the following be the two equations, in which $\scriptstyle {x}$ and $\scriptstyle {y}$ are the unknown quantities:—

$\scriptstyle {\left\{{\begin{aligned}&\scriptstyle {mx+ny=d}\\&\scriptstyle {m'x+n'y=d'.}\end{aligned}}\right.}$

We deduce $\scriptstyle {x={\frac {dn'-d'n}{n'm-nm'}}}$, and for $\scriptstyle {y}$ an analogous expression. Let us continue to represent by $\scriptstyle {\mathbf {V} _{0}}$, $\scriptstyle {\mathbf {V} _{1}}$, $\scriptstyle {\mathbf {V} _{2}}$, &c. the different columns which contain the numbers, and let us suppose that the first eight columns have been chosen for expressing on them the numbers represented by $\scriptstyle {m}$, $\scriptstyle {n}$, $\scriptstyle {d}$, $\scriptstyle {m'}$, $\scriptstyle {n'}$, $\scriptstyle {d'}$, $\scriptstyle {n}$ and $\scriptstyle {n'}$, which implies that $\scriptstyle {\mathbf {V} _{0}=m}$, $\scriptstyle {\mathbf {V} _{1}=n}$, $\scriptstyle {\mathbf {V} _{2}=d}$, $\scriptstyle {\mathbf {V} _{3}=m'}$, $\scriptstyle {\mathbf {V} _{4}=n'}$, $\scriptstyle {\mathbf {V} _{5}=d'}$, $\scriptstyle {\mathbf {V} _{6}=n}$, $\scriptstyle {\mathbf {V} _{7}=n'}$.

The series of operations commanded by the cards, and the results obtained, may be represented in the following table:—

Number of the operations.	Operation-cards.	Cards of the variables.		Progress of the operations.
	Symbols indicating the nature of the operations.	Columns on which operations are to be performed.	Columns which receive results of operations.
1	$\scriptstyle {\times }$	$\scriptstyle {\mathbf {V} _{2~}\times \mathbf {V} _{4~}=}$	$\scriptstyle {\mathbf {V} _{8~}\ldots \ldots \ldots }$	$\scriptstyle {=dn'}$
2	$\scriptstyle {\times }$	$\scriptstyle {\mathbf {V} _{5~}\times \mathbf {V} _{1~}=}$	$\scriptstyle {\mathbf {V} _{9~}\ldots \ldots \ldots }$	$\scriptstyle {=d'n}$
3	$\scriptstyle {\times }$	$\scriptstyle {\mathbf {V} _{4~}\times \mathbf {V} _{0~}=}$	$\scriptstyle {\mathbf {V} _{10}\ldots \ldots \ldots }$	$\scriptstyle {=n'm}$
4	$\scriptstyle {\times }$	$\scriptstyle {\mathbf {V} _{1~}\times \mathbf {V} _{3~}=}$	$\scriptstyle {\mathbf {V} _{11}\ldots \ldots \ldots }$	$\scriptstyle {=nm'}$
5	$\scriptstyle {-}$	$\scriptstyle {\mathbf {V} _{8~}-\mathbf {V} _{9~}=}$	$\scriptstyle {\mathbf {V} _{12}\ldots \ldots \ldots }$	$\scriptstyle {=dn'-d'n}$
6	$\scriptstyle {-}$	$\scriptstyle {\mathbf {V} _{10}-\mathbf {V} _{11}=}$	$\scriptstyle {\mathbf {V} _{13}\ldots \ldots \ldots }$	$\scriptstyle {=n'm-nm'}$
7	$\scriptstyle {\div }$	$\scriptstyle {{\frac {\mathbf {V} _{12}}{\mathbf {V} _{13}}}\quad =}$	$\scriptstyle {\mathbf {V} _{14}\ldots \ldots \ldots }$	$\scriptstyle {=x={\frac {dn'-d'n}{n'm-nm'}}}$

Since the cards do nothing but indicate in what manner and on what columns the machine shall act, it is clear that we must still, in every particular case, introduce the numerical data for the calculation. Thus, in the example we have selected, we must previously inscribe the numerical values of $\scriptstyle {m}$, $\scriptstyle {n}$, $\scriptstyle {d}$, $\scriptstyle {m'}$, $\scriptstyle {n'}$, $\scriptstyle {d'}$, in the order and on the columns indicated, after which the machine when put in action will give the value of the unknown quantity $\scriptstyle {x}$ for this particular case. To obtain the value of $\scriptstyle {y}$, another series of operations analogous to the preceding must be performed. But we see that they will be only four in number, since the denominator of the expression for $\scriptstyle {y}$, excepting the sign, is the same as that for $\scriptstyle {x}$, and equal to $\scriptstyle {n'm-nm'}$. In the preceding table it will be remarked that the column for operations indicates four successive multiplications, two subtractions, and