Question

write as a single fraction.\\(\frac{8y}{y^2 - 81} - \frac{1}{y - 2} - \frac{5}{y + 9}\\)simplify your answer as much as possible.

Explanation:

Step1: Factor the denominator

First, factor \(y^2 - 81\) using the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\). So \(y^2 - 81=(y + 9)(y - 9)\).
The expression becomes \(\frac{8y}{(y + 9)(y - 9)}-\frac{1}{y - 2}-\frac{5}{y + 9}\).

Step2: Find the common denominator

The common denominator of \((y + 9)(y - 9)\), \(y - 2\) and \(y + 9\) is \((y + 9)(y - 9)(y - 2)\).

Step3: Rewrite each fraction with the common denominator

• For \(\frac{8y}{(y + 9)(y - 9)}\), multiply numerator and denominator by \((y - 2)\): \(\frac{8y(y - 2)}{(y + 9)(y - 9)(y - 2)}\)
• For \(\frac{1}{y - 2}\), multiply numerator and denominator by \((y + 9)(y - 9)\): \(\frac{(y + 9)(y - 9)}{(y + 9)(y - 9)(y - 2)}\)
• For \(\frac{5}{y + 9}\), multiply numerator and denominator by \((y - 9)(y - 2)\): \(\frac{5(y - 9)(y - 2)}{(y + 9)(y - 9)(y - 2)}\)

Step4: Combine the fractions

\[

$$\begin{align*} &\frac{8y(y - 2)-(y + 9)(y - 9)-5(y - 9)(y - 2)}{(y + 9)(y - 9)(y - 2)}\\ =&\frac{8y^2-16y-(y^2 - 81)-5(y^2-11y + 18)}{(y + 9)(y - 9)(y - 2)}\\ =&\frac{8y^2-16y - y^2 + 81-5y^2 + 55y-90}{(y + 9)(y - 9)(y - 2)}\\ =&\frac{(8y^2 - y^2-5y^2)+(-16y + 55y)+(81 - 90)}{(y + 9)(y - 9)(y - 2)}\\ =&\frac{2y^2 + 39y-9}{(y + 9)(y - 9)(y - 2)} \end{align*}$$

\]

Answer:

\(\frac{2y^2 + 39y - 9}{(y + 9)(y - 9)(y - 2)}\)