Question

find the quotient.
\\(\frac{2x - 3}{x} div \frac{7}{x^2}\\)
\\(\bigcirc\\) a. \\(\frac{7}{x(2x - 3)}\\)
\\(\bigcirc\\) b. \\(\frac{7x}{2x - 3}\\)
\\(\bigcirc\\) c. \\(\frac{2x - 3}{7x}\\)
\\(\bigcirc\\) d. \\(\frac{x(2x - 3)}{7}\\)

Explanation:

Step1: Recall division of fractions

To divide two fractions, multiply the first fraction by the reciprocal of the second fraction. So, $\frac{2x - 3}{x} \div \frac{7}{x^2}$ becomes $\frac{2x - 3}{x} \times \frac{x^2}{7}$.

Step2: Multiply the numerators and denominators

Multiply the numerators: $(2x - 3) \times x^2 = x^2(2x - 3)$. Multiply the denominators: $x \times 7 = 7x$. Now we have $\frac{x^2(2x - 3)}{7x}$.

Step3: Simplify the fraction

Cancel out one $x$ from the numerator and the denominator. So, $\frac{x^2(2x - 3)}{7x}=\frac{x(2x - 3)}{7}$.

Answer:

D. $\frac{x(2x - 3)}{7}$