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Ch. R - Algebra Review
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161, 9780135440780, 9780136881117, 9780135440827, 9780135924891Non è quello che usi tu?Cambia libro di testo
Capitolo 1, Problema 49

Add or subtract, as indicated. See Example 4. (3/2k) + (5/3k)

Guida verificata passo dopo passo
1
Identify the given expression: \(\frac{3}{2k} + \frac{5}{3k}\).
Since the denominators are different, find the least common denominator (LCD). Here, the denominators are \$2k$ and \$3k$. The LCD is \$6k$ because \(6\) is the least common multiple of \(2\) and \(3\), and $k$ is common in both.
Rewrite each fraction with the LCD as the new denominator by multiplying numerator and denominator appropriately: - For \(\frac{3}{2k}\), multiply numerator and denominator by \(3\) to get \(\frac{3 \times 3}{2k \times 3} = \frac{9}{6k}\). - For \(\frac{5}{3k}\), multiply numerator and denominator by \(2\) to get \(\frac{5 \times 2}{3k \times 2} = \frac{10}{6k}\).
Now that both fractions have the same denominator, add the numerators: \(\frac{9}{6k} + \frac{10}{6k} = \frac{9 + 10}{6k} = \frac{19}{6k}\).
Express the final answer as a single simplified fraction: \(\frac{19}{6k}\). This is the sum of the original expression.

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