Question

topic: adding and subtracting rational expressions
progress:
the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer.
perform the indicated operations, then simplify.
\\(\frac{3b^2}{4a^2} + \frac{5b}{3a} - \frac{1}{6a^3}\\)
\\(\frac{3b^2 + 5b - 1}{4a^2 + 3a - 6a^3}\\)
\\(\frac{3b^2 + 5b - 1}{a^6}\\)
\\(\frac{3b^2 + 5b - 1}{12a^3}\\)
\\(\frac{20a^2b - 2 + 9ab^2}{12a^3}\\)

Explanation:

Step1: Find the LCD

The denominators are \(4a^2\), \(3a\), and \(6a^3\). Prime factorize each:

• \(4a^2 = 2^2 \cdot a^2\)
• \(3a = 3 \cdot a\)
• \(6a^3 = 2 \cdot 3 \cdot a^3\)

The LCD (Least Common Denominator) is the product of the highest powers of all prime factors: \(2^2 \cdot 3 \cdot a^3 = 12a^3\).

Step2: Rewrite each fraction with LCD

• For \(\frac{3b^2}{4a^2}\): Multiply numerator and denominator by \(3a\) (since \(12a^3 \div 4a^2 = 3a\)):

\(\frac{3b^2 \cdot 3a}{4a^2 \cdot 3a} = \frac{9ab^2}{12a^3}\)

• For \(\frac{5b}{3a}\): Multiply numerator and denominator by \(4a^2\) (since \(12a^3 \div 3a = 4a^2\)):

\(\frac{5b \cdot 4a^2}{3a \cdot 4a^2} = \frac{20a^2b}{12a^3}\)

• For \(\frac{1}{6a^3}\): Multiply numerator and denominator by \(2\) (since \(12a^3 \div 6a^3 = 2\)):

\(\frac{1 \cdot 2}{6a^3 \cdot 2} = \frac{2}{12a^3}\)

Step3: Perform the operations

Now, substitute back into the original expression:
\(\frac{9ab^2}{12a^3} + \frac{20a^2b}{12a^3} - \frac{2}{12a^3}\)

Combine the numerators over the common denominator:
\(\frac{9ab^2 + 20a^2b - 2}{12a^3}\)

Rearrange the numerator terms:
\(\frac{20a^2b - 2 + 9ab^2}{12a^3}\)

Answer:

\(\frac{20a^2b - 2 + 9ab^2}{12a^3}\) (corresponding to the last option)