Question

(t-5)/(1-t)-(t+4)/(t+1)+(t+2)/(t^(2)-1). Simplify using the LCD

Answer 1

To simplify the expression

\[
\frac{t-5}{1-t} - \frac{t+4}{t+1} + \frac{t+2}{t^2 - 1},
\]

we start by rewriting \(t^2 - 1\) because it can be factored into \((t-1)(t+1)\). The expression now reads:

\[
\frac{t-5}{1-t} - \frac{t+4}{t+1} + \frac{t+2}{(t-1)(t+1)}.
\]

Notice that \(1 - t = -(t - 1)\), so we can rewrite the first term as:

\[
\frac{t-5}{-(t-1)} = -\frac{t-5}{t-1}.
\]

Now the expression becomes:

\[
-\frac{t-5}{t-1} - \frac{t+4}{t+1} + \frac{t+2}{(t-1)(t+1)}.
\]

Next, we find the least common denominator (LCD) for all the terms. The denominators are \((t-1)\), \((t+1)\), and \((t-1)(t+1)\). Thus, the LCD is \((t-1)(t+1)\).

Now we convert each term to have a denominator of \((t-1)(t+1)\):

1. For the first term:

\[
-\frac{(t-5)(t+1)}{(t-1)(t+1)}.
\]

2. For the second term:

\[
- \frac{(t+4)(t-1)}{(t+1)(t-1)}.
\]

3. The third term is already:

\[
\frac{t+2}{(t-1)(t+1)}.
\]

Now we can rewrite the expression as:

\[
-\frac{(t-5)(t+1) + (t+4)(t-1) + (t+2)}{(t-1)(t+1)}.
\]

Next, we compute the numerator:

\[
-(t-5)(t+1) = -((t^2 + t - 5t - 5)) = -t^2 + 4t + 5,
\]

\[
-(t+4)(t-1) = -(t^2 - t + 4t - 4) = -t^2 + 3t + 4,
\]

So combining gives:

\[
(-t^2 + 4t + 5) + (-t^2 + 3t + 4) + (t + 2).
\]

Combine the like terms:

\[
-2t^2 + (4t + 3t + t) + (5 + 4 + 2) = -2t^2 + 8t + 11.
\]

Therefore, the final expression is:

\[
\frac{-2t^2 + 8t + 11}{(t-1)(t+1)}.
\]

Now we can simplify the final expression if possible, but \( -2t^2 + 8t + 11 \) doesn't have a simpler factorization.

So the final simplified expression is:

\[
\frac{-2t^2 + 8t + 11}{(t-1)(t+1)}.
\]