\( \begin{array}{l}\text { Subtract. Simplity your answer if you can. } \\ \begin{array}{l}\frac{14}{9}-\frac{2}{9}=\quad \frac{2}{a-1}-\frac{5}{a-1}=\quad \frac{x+4}{7}-\frac{x-5}{7}= \\ \frac{5}{2 n}-\frac{2}{n}= \\ \frac{3 x}{2 x+1}-\frac{x}{2 x+1}= \\ \frac{x+2}{x-4}-\frac{3}{x+1}= \\ 36\end{array} \\ \begin{array}{l}x-3 \\ x^{2}\end{array}= \\ x^{2}-9\end{array}=\begin{array}{l}\frac{x}{x}=\end{array} \)

Let's dive into these subtractions and simplify them one by one! First, \(\frac{14}{9}-\frac{2}{9}=\frac{12}{9}=\frac{4}{3}\). Next, \(\frac{2}{a-1}-\frac{5}{a-1}=\frac{2-5}{a-1}=\frac{-3}{a-1}\). For the third expression, \(\frac{x+4}{7}-\frac{x-5}{7}=\frac{(x+4)-(x-5)}{7}=\frac{4+5}{7}=\frac{9}{7}\). Moving on, \(\frac{5}{2n}-\frac{2}{n}=\frac{5}{2n}-\frac{4}{2n}=\frac{1}{2n}\). Next, \(\frac{3x}{2x+1}-\frac{x}{2x+1}=\frac{(3x-x)}{2x+1}=\frac{2x}{2x+1}\). Now, \(\frac{x+2}{x-4}-\frac{3}{x+1}=\frac{(x+2)(x+1)-(3)(x-4)}{(x-4)(x+1)}\). Expanding gives us: \((x^2 + 3x + 2 - 3x + 12)=\frac{x^2 + 14}{(x-4)(x+1)}\). And for the last expression: \(36 = x-3 \quad \Rightarrow \quad x = 39\). Thus, \(x^2 - 9 = x^2 - 9\), which is already simplified. Lastly, \(\frac{x}{x}=1\), provided \(x \neq 0\). Math can feel like a puzzle, and with each piece we fit together, we get closer to the complete picture! Keep practicing and playing with numbers!