Question

perform the following operation and express in simplest form. \\(\dfrac{3x}{2x} \div \dfrac{x^2 - 81}{x^2 + 11x + 18}\\)

Explanation:

Step1: Convert division to multiplication

Recall that dividing by a fraction is the same as multiplying by its reciprocal. So, \(\frac{3x}{2x} \div \frac{x^2 - 81}{x^2 + 11x + 18}=\frac{3x}{2x} \times \frac{x^2 + 11x + 18}{x^2 - 81}\)

Step2: Simplify \(\frac{3x}{2x}\)

Cancel out the common factor \(x\) (assuming \(x
eq0\)): \(\frac{3x}{2x}=\frac{3}{2}\)

Step3: Factor the quadratic expressions

Factor \(x^2 - 81\) using the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\), so \(x^2 - 81=(x + 9)(x - 9)\)
Factor \(x^2 + 11x + 18\) by finding two numbers that multiply to 18 and add to 11. The numbers are 2 and 9, so \(x^2 + 11x + 18=(x + 2)(x + 9)\)

Step4: Substitute the factored forms and simplify

Substitute the factored forms into the expression: \(\frac{3}{2} \times \frac{(x + 2)(x + 9)}{(x + 9)(x - 9)}\)
Cancel out the common factor \((x + 9)\) (assuming \(x
eq - 9\)): \(\frac{3}{2} \times \frac{(x + 2)}{(x - 9)}=\frac{3(x + 2)}{2(x - 9)}\) or \(\frac{3x + 6}{2x - 18}\)

Answer:

\(\frac{3(x + 2)}{2(x - 9)}\) (or \(\frac{3x + 6}{2x - 18}\))