As division is the reverse of multiplication, so to divide one quantity by another is but to retrace the steps taken in multiplication. If we have the area ab (Fig. 278), and one of the factors a given to find the other, we have but to remove from a b the factor a, and write the answer b.
If we have the cubic contents of a solid abc (Fig. 279). and one of the factors a given to find the area represented by the other two, we have but to remove a, and write the others, b c, as the answer.
If there be given the area represented by a (b + c) (see Art. 411), and one of the factors a to find the other, we have but to remove a and write the answer b + c. Sometimes, however, a (b + c) is written a b + ac. Then the given factor is to be removed from each monomial and the answer written b + c.
If there be given the area represented by a2 + 2 a b + b2 to find the factors, then we know by Art. 412 that this area is that of a square the sides of which measure a + b, and that the area is the product of a + b by a + b; or, that a + b is the square root of a2 + 2 a b + b2.
If there be given the area a2 - 2 ab + b2 to find its factors, then we know by Art. 413 that this area is that of a square whose sides measure a - b, or that it is the product of a - b by a - b, or the square of a - b.
If there be given the difference, of the squares of two quantities, or the area represented by a2 - b2, to find its factors, then we know by Art. 414 that this is the area produced by the multiplication of a - b by a + b.
4-18- - Division: Statement of Quotient. - In any case of division the requirement may be represented as a fraction; thus: To divide c + d - f by a - b we write the quotient thus -
c + d - f a - b
For example, to illustrate by numerals, let a = 7, b = 3, c = 4, d = 5, and f == 6. Then the above becomes -
4 + 5 - 6 = 3
7- 3 4
419, - Division; Reduction. - When each monomial in either the numerator or denominator contains a common quantity, that quantity may be removed and placed outside of parentheses containing the monomials from which it was taken; thus, in -
2 ab + 4 ac - 8 ad
we have 2 and a factors common to each monomial of the numerator. Therefore the expression may be reduced to
To test this arithmetically we willl put a = 9, b = 7, c = $, d = 4, and / = 6. Then for the first expression we have -
which equals -
126 + 180 - 288 6 = 3
And for the second expression -
which equals -
18 (17- 16) = 18 = 3;
the same result as before. It will be observed that in this process of removing all common factors algebra furnishes the means of performing the work arithmetically with many less figures. The reduction is greater when the common factors are found in both numerator and denominator. For example, in the expression -
12dn - 18fn
we have 3 n a factor common to each monomial in the numerator and denominator; therefore the expression reduces to
3n(a + 3b - 5 c) 3n (4d - 6f)
And now, since 3 n is a factor common to both numerator and denominator, these cancel each other; therefore (Art. 371) the expression reduces to -
a + 3b - 5 c 4d- 6f
To test these reductions arithmetically, let a = 9, b - 8, c - 4, d - 6, / = 3, and n = 5. Then the first expression becomes -
which equals -
135 + 360-300 = 195 = 2 1 360 - 270 90 6
and the second expression becomes -
which equals -
9 + 24 - 20 = 13 = 2 1/6
The same result, but with many less figures.
420. - Proportional: Analysis. - In the formula of the lever (Art, 377), P x CF - R x E C. Let n be put for the arm of leverage CF and m for E C. Then we have -
Pn = R m, from which by division (Art. 372) we have (Art. 399) -
R = Pn/m, (111.)
Suppose there be a case in which neither R nor P severally are known, but that their sum is known; and it is required from this and the m and n to find R and P. Let -
W= R + P,
then - W - R = P. (See Art. 403.)
The value of P was above found to be -
P= R m/n
Since P= R m/n - and also equals W - R, therefore -
W -R = R m/n.
Transferring R to the opposite member (Art. 403) we have -
W=R + R - m/n
Here R appears as a common factor and may be separated by division (Art. 419); thus -
w=r (1 + m/n).
By division the factor ( 1 + m/n) may be transferred to the
opposite member (Art. 371). Thus we have -
R= W/1 + m/n, (116.)
by which we find the value of R developed. As an example, let W = 1000 pounds, m = 3 feet and n = 7 feet; then -
R = 1000/1 + 3/7 = 1000/10/7. Multiplying the numerator and denominator by 7, we get -
R = 7x1000/10 = 700.
R + P
But a process similar to the above develops an expression for the value of P, which is -
P=W/1 + n/m (117.)
Putting this to the test of figures, we have -
P = 1000/1 + 7/3 = 1000/10/3 = 3000/10 = 300.