Let it be required to add the fractions 1/5 and 2/5. By referring to Art. 379 we see that A D (Fig. 274), is one of the five parts into which the whole line A B is divided; it is, therefore, 1/5 We also see that D C contains two of the five parts; it is, therefore,2/5. We also see that A D + D C= A C, which contains three of the five parts, or A C = 3/5 of A B

We therfore conclude that 1/5+2/5=3/5. In this operation it is seen that the denominator is not changed, and that the resultant fraction has for a numerator a number equal to the sum of the numerators of the fractions which were required to be added.

By this it is shown that to add fractions we simply take the stun of the numerators for the new numerator, making the denominator of the resultant fraction the same as that of the fractions to be added. For example: What is the sum of the fractions 1/9, 3/9 and 4/5? Here we have 1+3+4 = 8 for the numerator, therefore -

1/9= 3/9= 4/9 =8/ 9