Elements of Algebra

CHAP. IX. OF ALGEBRA. 27 4. Reduce $\frac{x^2-y^2}{x^4-y^4}$ to its lowest terms. Ans. $\frac{1}{x^2+y^2}$. 5. Reduce $\frac{a^4-x^4}{a^3-a^2x-ax^2+x^3}$ to its lowest terms. Ans. $\frac{a^2+x^2}{a-x}$. 6. Reduce $\frac{5a^5+10a^4x+5a^3x^2}{a^3x+2a^2x^2+2ax^3+x^4}$ to its lowest terms. Ans. $\frac{5a^4+5a^3x}{a^2x+ax^2+x^3}$. *** CHAP. IX. Of the Addition and Subtraction of Fractions. 94. When fractions have equal denominators, there is no difficulty in adding and subtracting them; for $\frac{2}{7} + \frac{3}{7}$ is equal to $\frac{5}{7}$, and $\frac{4}{7} - \frac{2}{7}$ is equal to $\frac{2}{7}$. In this case, therefore, either for addition or subtraction, we alter only the numerators, and place the common denominator under the line, thus; $\frac{7}{100} + \frac{9}{100} - \frac{12}{100} - \frac{15}{100} + \frac{20}{100}$ is equal to $\frac{9}{100}$; $\frac{24}{50} - \frac{7}{50} - \frac{12}{50} + \frac{31}{50}$ is equal to $\frac{36}{50}$, or $\frac{18}{25}$; $\frac{16}{20} - \frac{3}{20} - \frac{11}{20} + \frac{14}{20}$ is equal to $\frac{16}{20}$, or $\frac{4}{5}$; also $\frac{1}{3} + \frac{2}{3}$ is equal to $\frac{3}{3}$, or 1, that is to say, an integer; and $\frac{2}{4} - \frac{3}{4} + \frac{1}{4}$ is equal to $\frac{0}{4}$, that is to say, nothing, or 0. 95. But when fractions have not equal denominators, we can always change them into other fractions that have the same denominator. For example, when it is proposed to add together the fractions $\frac{1}{2}$ and $\frac{1}{3}$, we must consider that $\frac{1}{2}$ is the same as $\frac{3}{6}$, and that $\frac{1}{3}$ is equivalent to $\frac{2}{6}$; we have therefore, instead of the two fractions proposed, $\frac{3}{6} + \frac{2}{6}$, the sum of which is $\frac{5}{6}$. And if the two fractions were united by the sign minus, as $\frac{1}{2} - \frac{1}{3}$, we should have $\frac{3}{6} - \frac{2}{6}$, or $\frac{1}{6}$. As another example, let the fractions proposed be $\frac{3}{4} + \frac{5}{8}$. Here, since $\frac{3}{4}$ is the same as $\frac{6}{8}$, this value may be substituted for $\frac{3}{4}$, and we may then say $\frac{6}{8} + \frac{5}{8}$ makes $\frac{11}{8}$, or $1\frac{3}{8}$. Suppose farther, that the sum of $\frac{1}{3}$ and $\frac{1}{4}$ were required, I say that it is $\frac{7}{12}$; for $\frac{1}{3} = \frac{4}{12}$, and $\frac{1}{4} = \frac{3}{12}$: therefore $\frac{1}{3} + \frac{1}{4} = \frac{7}{12}$. 96. We may have a greater number of fractions to reduce