EQUATIONS OF THE FIRST DEGREE, CONTAINING BUT ONE UNKNOWN QUANTITY.

All the operations, that are used to solve an equation of the first degree, with but one unknown quantity, that is, to have this unknown quantity by itself in the first member free from every coefficient, are reduced to four, Addition, Subtraction, Multiplication, and Division; and we operate from this evident principle, that the equality existing between two quantities, is not destroyed by adding to, subtracting from, multiplying and dividing each of them by the same quantity.

The method for solving an equation is always to begin with clearing the unknown quantity of its denominators, if it has any; then all the terms that contain the unknown quantity, are placed in the first member of the equation, and all the known quantities in the second; finally, the second member is divided by the coeffi cient of the unknown quantity, which is then left by itself in the first member.

We will now apply these principles in different examples.

Be the equation x + a → bd.

In order to solve this equation, we must get x by itself in the first member. In the first place, I re

m

Such an equation as a, may be confidered as one of the first degree; in order to folve it, both members must be raised to the power m, which will give xam; if ab was the equation, it might also be confidered as one of the first dez gree, and folved by the affiffance of logarithms.

mark, that if the quantity b is added to each member of the equation, it will disappear from the first member; for then the equation, will be x + a- -6+6 = d+ b, which is reduced to x + a = d + b.

Again, I remark that by subtracting a from each member of the equation, this quantity will disappear from the first member; for then the equation will be, x+a~a=d+b—a, and à =d+b−a. Here, then, is the solution of the equation; and if determined values were given to a, b, d, we should have a determined value for x.

If we reflect on the operations just performed, we may draw from them general rules. In order to disengage x, we see that addition and subtraction have been successively made use of; the principle of preserving the equality of the two quantities being constantly observed, by adding to the one what we added to the other, and subtracting from the one what we subtracted from the other.

But we see also, by the final equation x=d+b-11 that the result of those two operations is reduced to have a and b in the second member of the equation, with signs directly contrary to those they had in the 1st member; as the reasoning, by which we obtained this result, is not confined particularly to a and b, we conclude generally, that in order to clear a member of an equation of any term, that term must be placed in the other member with a sign contrary to the one it had, This rule affords also a simple way of changing a term from one member to the other, which is called transpo sition.

Let 36+ be a given equation.

In order to resolve this equation, all the terms in which x is found, must be transposed into the first member, and all the known quantities into the other; thus we shall have 3 x-x—6 + a, and 2 x➡b +a; b + a

finally x

2

Let the given equation be pɑ = b + *,

We shall have px - · x = b+a, and (p−1 ) x=b+4,

b + a

paid to this manner of clearing the unknown quantity, when it is found in several terms of the equation, with different coefficients. All the different coefficients of the unknown quantity should be united under the same parenthesis, as was just now done; then this assemblage should be considered as one coefficient. This process is by no means difficult to understand, for on performing the multiplication as it is expressed, it is evident, we shall have the same terms we had before this change of form.

up

With regard to the sign of each of the coefficients, comprised between the two same parentheses to form one coefficient, we must remark, that each of these coefficients must preserve the sign it had before, if the sign +be before the parenthesis, and on the contrary if it be the sign, the sign of each coefficient must be changed.

An example will shew the necessity of this change. Let it be ab xc x=d,

In order to assemble the coefficients of x we must write ab+c)x=d; for, upon performing the indicated multiplication, and observing the rule for signs demonstrated under the head of fractions, we shall have the equation under its original form in such case beginners are apt to mistake the signs.

I continue solving the equation, and I have —(b+c)

2

x=d-a; from which I draw.

- x=

d-a2

b + c

changing the signs in both members, I shall have

a2-d

b+c

Take notice here, that the signs of the denominator have not been changed, the reason of which has been already given.

b d

b-x

Take the equation

a c b

C

According to the method laid down, we must begin with clearing the unknown quantity of its denominators; the most simple way of doing it, is, in the first place, to reduce to a common denominator the terms of each member separately, and then do the same by both members.

We shall have in the first place,

cx-ab cd-b2+bx

ac

bc

and as c is common to both the denominators, noth

ing more is necessary, than to multiply the first mem

ber by b, and the second by a, which givesacd-ab2 + a b x

a b c

bcx-ab

a b c

Now, it is very evident, that

we can take away the denominators; for here are two equal fractions with the same denominator, consequently the numerators must be equal; thus we shall have b c x ab2acd- a b2 + a b x ; let it be remarked, that in the two members, there are two like terms with the same signs, which may therefore be thrown away without injuring the equality: Then the equation is reduced to bexacd+abx, and by transposition bcx-abxaed, or (bcab) x =

acd, from which, finally, a =

acd

bc-ab

Let us propose again to solve this equation,

a + x

b.

d

Before proceeding to clear away the denominators, the first term of the equation requires some preparation; its denominator, which is composed of a whole quantity and a fraction, should be entirely reduced

under the fractional form, and then it becomes

CX

C

b

;

;

thus we have, therefore, a + a to'divide by a fraction that is to say, according to a principle of arithmetic, this quantity must be multiplied by the denominator of the fraction, and the product must be divided by