Question

which shows the expressions rewritten with the least common denominator?\\(\frac{7x - 2}{4x^2}\\) and \\(\frac{x - 1}{8x}\\)\\(\frac{28x - 8}{16x^2}\\) and \\(\frac{2x^2 - 2x}{16x^2}\\)\\(\frac{14x - 4}{8x^2}\\) and \\(\frac{2x - 2}{8x^2}\\)\\(\frac{28x^2 - 8x}{16x^3}\\) and \\(\frac{2x^3 - 2x^2}{16x^3}\\)\\(\frac{14x - 4}{8x^2}\\) and \\(\frac{x^2 - x}{8x^2}\\)

Explanation:

Step1: Find the least common denominator (LCD) of \(4x^2\) and \(8x\).

Factor the denominators: \(4x^2 = 2^2 \cdot x^2\), \(8x = 2^3 \cdot x\). The LCD is the product of the highest powers of all prime factors, so \(2^3 \cdot x^2 = 8x^2\).

Step2: Rewrite \(\frac{7x - 2}{4x^2}\) with LCD \(8x^2\).

Multiply numerator and denominator by 2: \(\frac{(7x - 2) \cdot 2}{4x^2 \cdot 2} = \frac{14x - 4}{8x^2}\).

Step3: Rewrite \(\frac{x - 1}{8x}\) with LCD \(8x^2\).

Multiply numerator and denominator by \(x\): \(\frac{(x - 1) \cdot x}{8x \cdot x} = \frac{x^2 - x}{8x^2}\).

Answer:

\(\frac{14x - 4}{8x^2}\) and \(\frac{x^2 - x}{8x^2}\) (the fourth option)