Case IV.

131. To Reduce Fractions to their Lowest Common Denominator.

The process is the same as in Arithmetic. Hence:

Find the lowest common multiple of the denominators; this will be the required denominator. Divide this denominator by the denominator of each fraction.

Multiply the first numerator by the first quotient, the second numerator by the second quotient, and so on.

The products will be the respective numerators of the equivalent fractions.

Every fraction should be in its lowest terms before the common denominator is found.

1. Reduce ${\displaystyle \frac{3x}{4a^2}}$, ${\displaystyle \frac{2y}{3a}}$, and ${\displaystyle \frac{5}{6a^3}}$ to equivalent fractions having the lowest common denominator.

The L. C. M. of $4a^2$, $3a$, and $6a^3=12a^3$.

The respective quotients are $3a$, $4a^2$, and $2$.

The products are $9ax$, $8a^{2}y$, and $10$.

Hence, the required fractions are $\frac{9ax}{12a^3},\frac{8a^{2}y}{12a^3},\text{and}\frac{10}{12a^3}.$

2. Express ${\displaystyle \frac{1}{x^2+5x+6}}$ and ${\displaystyle \frac{1}{x^2+4x+3}}$ with lowest common denominator.

The factors of the denominators are $x+3$, $x+2$; and $x+3$, $x+1$.

Hence the lowest common denominator (L. C. D.) is $(x+3)(x+2)(x+1)$, and the required numerators are $x+1$ and $x+2$. Hence the required fractions are $\frac{x+1}{(x+3)(x+2)(x+1)}\text{and}\frac{x+2}{(x+3)(x+2)(x+1)}.$

Exercise 45.

Express with lowest common denominator:

1. ${\displaystyle \frac{x}{x-a}}$, ${\displaystyle \frac{x^2}{x^2-a^2}}$.

2. ${\displaystyle \frac{a}{a+b}}$, ${\displaystyle \frac{a^2}{a^2-b^2}}$.

3. ${\displaystyle \frac{1}{1+2a}}$, ${\displaystyle \frac{1}{1-4a^2}}$.

4. ${\displaystyle \frac{9}{16-x^2}}$, ${\displaystyle \frac{4-x}{4+x}}$.