Lesson 3 · from §63 of the 1910 text
Equivalent Fractions and the L.C.D.
Two fractions are equivalent when you can get one from the other by multiplying or dividing both terms by the same expression: \frac{2}{3} and \frac{8}{12} are equivalent, and so are \frac{a^2}{ab} and \frac{a}{b}.
The lowest common denominator (L.C.D.) of some fractions is the L.C.M. of their denominators, which you learned to find in the last chapter.
Example. Change \dfrac{5a}{6b^2c} and \dfrac{3b}{4a^2c^2} to equivalent fractions with the L.C.D. The L.C.M. of the denominators is 12a^2b^2c^2. The first denominator is missing 2a^2c, so multiply both its terms by that: \dfrac{10a^3c}{12a^2b^2c^2}. The second is missing 3b^2: \dfrac{9b^3}{12a^2b^2c^2}.
Example. \dfrac{3x - 1}{2x^3 - 18x} and \dfrac{5}{x^2 - 5x + 6}. Factor: 2x(x + 3)(x - 3) and (x - 3)(x - 2). The L.C.D. is 2x(x + 3)(x - 3)(x - 2). Multiply the first fraction's terms by x - 2 and the second's by 2x(x + 3): \frac{(3x - 1)(x - 2)}{2x(x + 3)(x - 3)(x - 2)}, \qquad \frac{10x(x + 3)}{2x(x + 3)(x - 3)(x - 2)}.
Rule. Write the denominators in factored form and find their L.C.M. Then multiply the numerator and denominator of each fraction by the factors of the L.C.M. that are missing from its own denominator.
Typing tip. Type each new fraction with the L.C.D. underneath. The denominator can be factored or multiplied out, so (3x^2 - 7x + 2)/(2x(x+3)(x-3)(x-2)) is fine.
Quick check
\dfrac{3}{4x} and \dfrac{5}{6x^2}
From the history of algebra
Fractions were hard for every early civilization. The Egyptians, and the Greeks right up to the 500s AD, wrote fractions as sums of fractions with numerator 1: \frac58 was written \frac12 + \frac18. The Romans put everything over 12, and the Babylonians used 60 (which is why an hour has 60 minutes). Hindu mathematicians seem to have been the first to bring fractions to a common denominator the way we do, though Euclid (300 BC) already knew how to find the least common multiple of numbers.
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
Change each set of fractions to equivalent fractions with the lowest common denominator. You may type the denominator factored, like (5x+1)/(6(x+1)).
\dfrac{1}{2}, \dfrac{1}{3} and \dfrac{1}{6}
\dfrac{5}{24} and \dfrac{7}{32}
\dfrac{4}{3d} and \dfrac{2c}{5d}
\dfrac{3a+b}{12} and \dfrac{3a}{16}
\dfrac{5x+y}{21} and \dfrac{4x+3y}{35}
\dfrac{4}{5a^2b} and \dfrac{3}{abc}
\dfrac{5c}{3mn^3} and \dfrac{12d}{11m^2n^2}
\dfrac{3a}{2b^2c}, \dfrac{2b}{3a^2c} and \dfrac{c}{6ab}
\dfrac{3b}{2cd^3}, \dfrac{9a}{4cde} and \dfrac{7ab}{5cde^2}
\dfrac{x+y}{3xy^2} and \dfrac{x-2y}{2x^2y}
\dfrac{2+m^2}{2m} and \dfrac{4+m^2}{4n}
\dfrac{2}{x+2} and \dfrac{4}{3x+6}
\dfrac{4}{x+1} and \dfrac{5}{x-1}
\dfrac{3c}{c^2-d^2} and \dfrac{4}{c+d}
\dfrac{3x}{x+3} and \dfrac{5}{x^2+5x+6}
\dfrac{2x}{x^2-xy} and \dfrac{3y}{x^2-2xy+y^2}
\dfrac{3c+2}{c^2-d^2} and \dfrac{4-c}{c^2-7cd+6d^2}
\dfrac{2x+5}{x^2-1} and \dfrac{2x-4}{x^3-3x^2+2x}