It will be seen from the above that in order to solve two simultaneous equations of the first degree, we first deduce from the given equations a third equation which contains only one of the unknown quantities; and the unknown quantity which is absent is said to have been eliminated.

143. From the last article it will be seen that the values of x and y which satisfy the equations

we have

y

1

=

=

bc-b'c ca'-c'a ab'-a'b'

It is important that the student should be able to

quote these formulae.

These may be considered as two simultaneous equations of the

Instead of referring to the general formula of Art. 143, as we have done in the above examples, the unknown quantities may be eliminated in turn, as in Art. 142; and this latter method is frequently the simpler of the two. Thus in this last example we have at once, by multiplying the first equation by a and then subtracting the second,

144. Discussion of solution of two simultaneous equations of the first degree. We have seen that the values of x and y which satisfy the equations

(ba' - b'a) y = ca'-c'a......(iv).

Thus there is a single finite value of x, and a single finite value of y, provided that ab' — a'b0.

If ab' - a'b = 0, x will be infinite [see Art. 118] unless cb'-c'b0; and, if ab' - a'b and cb'-c'b are both zero, any value of x will satisfy equation (iii).

So also, y will be infinite if ab' - a'b=0, unless ca' - c'a is also zero, in which case any value of y will satisfy equation (iv).

When equations cannot be satisfied by finite values of the unknown quantities, they are often said to be inconsistent. Thus the equations ax + by = c and ax + b'y = c′ are inconsistent if

a

α

b

=

b

unless each fraction is equal to

in which case the equations are indeterminate. In fact

the two given equations are equivalent to one only.

We have hitherto supposed that a, a, b, b' were none of them zero. It will not be necessary to discuss every possible case: consider, for example, the case in which a and a' are both zero.

When a and a' are both zero, we have from (i) y

=

C

These results are inconsistent with

c'

(ii) are satisfied by making y=, and by giving to a any

finite value whatever.

C

If however + the equations by =c and b'y = c

b

cannot both be satisfied, unless they are looked upon as the limiting forms of the equations ax + by = c and a'x + b'y = c', in which a and a' are indefinitely small and ultimately zero. But from (iii) we see that when a and a' diminish. without limit, x must increase without limit, cb' — c'b not being zero. Thus, in the equations (i) and (ii), when a and a diminish without limit, and cb'c'b, the value of x must be infinite.

Equations with three unknown quantities.

145. To solve the three equations:

Method of successive elimination. Multiply the first equation by c', and the second by c; then we have

ac'x+bc'y + cc'z = dc',

and

a'cx+b'cy + c'cz = d'c;

therefore, by subtraction,

.....(iv).

(ac' — a'c) x + (bc' — b'c) y = dc' — d'c....... Again, by multiplying the first equation by c" and the third by c and subtracting, we have

(ac” — a′′c) x + (bc′′ — b′′c) y = dc" — d'c.. .(v).

......

We now have the two equations (iv) and (v) from which to determine the unknown quantities x and y. Using the general formulae of Art. 143, we have

X =

– (bc′ — b’c) (dc′′ — d'c) + (dc' — d'c) (bc'' — b′′c)

(ac′ — a'c) (bc′ — b′′c) — (bc' — b’c) (ac′′ — a′′c)

Method of undetermined multipliers. Multiply the equations (i) and (ii) by λ and μ, and add to (iii); then we have the equation

x (λa + μa' + a') + y (λb + μb′ + b′′) + z (λc + μc′ +c′′)

which is true for all values of λ and μ.

Now let λ and μ be so chosen that the co-efficients of y and z may both be zero,

X=

λ

λc + μc' + c = 0;

b'c"-b"c

=

=

1

b'c-bc" bc' - b'c'

d (b'c" — b′′c') + d' (b'c — bc") + d" (bc' — b'c)

a (b'c" — b′′ c') + a′ (b′′c — bc′) + a′′ (bc′ – b′c) '

The numerator and the denominator of the first value of x, which was obtained by eliminating z and y in succes