This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

To exemplify the subtraction of fractions, let it be required to find the algebraic sum of a/c-b/d-e/f. These denominators all differ. The fractions, therefore, require to be modified, so that each denominator shall contain them all. To accomplish this, the first fraction will need to be thus treated:

ax df=adf

cx df =cdf the second -

-b x cf = - b c f

d x cf = c df ; the third -

- e x cd = - cde

f x cd = cdf. The sum of these is -

a df - bcf - cde

cdf .

That this is a correct answer, let the result be proved by figures; thus, for a put 15; b, 2; c, 3; d, 4; e 5; f, 6. Then* we shall have -

a/c - b/d - e/f = 15/3 - 2/5 - 5/6

It will be observed that these denominators may be equalized by multiplying the first fraction by 2, and the second by 1 1/2, therefore we have -

30/6 - 3/6 - 5/6.

To make the required subtraction we are to deduct from 30 (the numerator of the positive fraction), first 3, then 5; or, the sum of the numerators of the negative fractions; or for the numerator of the new fraction we have 30 - 8 = 22. The required result, therefore, is -

22/6 = 11/3 = 3 2/3. To apply this test to the algebraic sum we have -

which by multiplication reduces to -

360 - 36 - 60 264 = 22 = 11 = 3 2/3,

72 72 6 3

a result the same as before, proving the work correct. Another example:

From a/n-b/m take c/n, d/m and e/n;

or. find the algebraic sum of -

a/n-b/m - c/n, d/m -e/n;

The fractions which have the same denominator may be grouped together thus:

a/n - c/d - e/n = a-c-e/n;

and -

b/m - d/m = b-d/m.

To harmonize these two denominators, m and n, the first fraction must be multiplied by m and the last by n, or -

m(a - c - e)/mn + n (b - d)/mn = m (a - c - e) + n(b - d)/mn In the polynomial factor within the parentheses (a - c - e) we have the positive quantity a, from which is to be taken the two negatives c and e, or their sum is to be taken from a, or (a - (c + e) ). With this modification we have for the algebraic sum of the five given fractions -

m(a - (c+e))+n(b - d)/mn. To test the accuracy of this result, let the value of the several letters respectively be as follows: a = 11,b = 9,c = 3,

d = 4, e = 5, m = 10, and n = 8. Then the sum is -

10 (11-(3+5) + 8(9-4)/10x8 = 70/80 = 7/8.

Now, taking the fractions separately, we have -

a/n-c/n-e/n=11/8-(3/8+5/8)=11/8-8/8=3/8;

again, - b/m - d/m = 9/10 - 4/10 = 5/10;

or, together we have, as the sum of these two results -

3/8 + 5/10.

To harmonize these denominators we may multiply the first fraction by 5, and the second by 4, thus:

3 x 5 = 15 , 5 x 4 = 20 8 x 5 = 40' 10x4 = 40'

and then the sum is -

15/40 + 20/40 = 35/40 = 7/8; the same result as before, thus the accuracy of the work is established.

Continue to: