Question

solve for t. if there are multiple solutions, enter them as a list of values separated by commas. if there are no solutions, type none.\\(\frac{8}{t - 2} + \frac{1}{t + 2} = \frac{1}{t^2 - 4}\\)\\(t = \\)\
enter any extraneous solutions. if there are multiple extraneous solutions, enter them as a list separated by a comma. if there are no extraneous solutions, enter none.

Explanation:

Step1: Factor the denominator

Notice that \(t^2 - 4=(t - 2)(t + 2)\), so the equation \(\frac{8}{t - 2}+\frac{1}{t + 2}=\frac{1}{t^2 - 4}\) can be rewritten as \(\frac{8}{t - 2}+\frac{1}{t + 2}=\frac{1}{(t - 2)(t + 2)}\).

Step2: Multiply through by the LCD

The least common denominator (LCD) of the fractions is \((t - 2)(t + 2)\). Multiply each term by \((t - 2)(t + 2)\) to eliminate the fractions:
\[ \begin{align*} 8(t + 2)+1(t - 2)&=1\\ 8t+16+t - 2&=1\\ 9t+14&=1 \end{align*} \]

Step3: Solve the linear equation

Subtract 14 from both sides: \(9t=1 - 14=-13\). Then divide by 9: \(t =-\frac{13}{9}\).

Step4: Check for extraneous solutions

We need to check if \(t =-\frac{13}{9}\) makes the original denominators zero. The denominators are \(t - 2\) and \(t + 2\).
- For \(t - 2\): \(-\frac{13}{9}-2=-\frac{13}{9}-\frac{18}{9}=-\frac{31}{9} eq0\).
- For \(t + 2\): \(-\frac{13}{9}+2=-\frac{13}{9}+\frac{18}{9}=\frac{5}{9} eq0\). Wait, but wait, maybe I made a mistake? Wait, no, let's re - check the algebra. Wait, when we multiplied through, let's redo step 2:

Wait, \(\frac{8}{t - 2}\times(t - 2)(t + 2)=8(t + 2)\), \(\frac{1}{t + 2}\times(t - 2)(t + 2)=(t - 2)\), and \(\frac{1}{(t - 2)(t + 2)}\times(t - 2)(t + 2)=1\). So the equation after multiplying is \(8(t + 2)+(t - 2)=1\), which is \(8t+16+t - 2 = 1\), \(9t + 14=1\), \(9t=-13\), \(t =-\frac{13}{9}\). But wait, let's plug \(t =-\frac{13}{9}\) back into the original equation:

Left - hand side (LHS): \(\frac{8}{-\frac{13}{9}-2}+\frac{1}{-\frac{13}{9}+2}=\frac{8}{-\frac{31}{9}}+\frac{1}{\frac{5}{9}}=8\times(-\frac{9}{31})+\frac{9}{5}=-\frac{72}{31}+\frac{9}{5}=\frac{- 360+279}{155}=\frac{-81}{155}\)

Right - hand side (RHS): \(\frac{1}{(-\frac{13}{9})^2-4}=\frac{1}{\frac{169}{81}-4}=\frac{1}{\frac{169 - 324}{81}}=\frac{1}{-\frac{155}{81}}=-\frac{81}{155}\)

Wait, so \(t =-\frac{13}{9}\) is a valid solution? But the problem asks for extraneous solutions. Wait, maybe I messed up the initial problem. Wait, no, the question first asks to solve for \(t\), then asks for extraneous solutions. Wait, maybe there was a miscalculation. Wait, let's start over.

Wait, the original equation: \(\frac{8}{t - 2}+\frac{1}{t + 2}=\frac{1}{t^2 - 4}\). The domain of the original function is \(t eq2\) and \(t eq - 2\) because those values make the denominators zero.

When we solve \(8(t + 2)+(t - 2)=1\), we get \(8t+16+t - 2 = 1\), \(9t+14 = 1\), \(9t=-13\), \(t =-\frac{13}{9}\), which is not equal to 2 or - 2, so it is a valid solution. But the problem's first part asks for \(t\) (the solution), and the second part asks for extraneous solutions. Since the solution we found is valid, there are no extraneous solutions. So the answer for extraneous solutions is None. And the solution for \(t\) is \(-\frac{13}{9}\), but wait, maybe the user made a mistake in the problem presentation? Wait, no, the user's problem first asks to solve for \(t\), then asks for extraneous solutions. Wait, maybe I misread the problem. Wait, the user's problem:

First: "Solve for \(t\). If there are multiple solutions, enter them as a list of values separated by commas. If there are no solutions, type None."

Second: "Enter any extraneous solutions... If there are no extraneous solutions, enter None."

Wait, when we solved, we got \(t =-\frac{13}{9}\), which is a valid solution. So the first answer (for \(t\)) is \(-\frac{13}{9}\), and the second answer (for extraneous solutions) is None. But maybe there was a mistake in my first calculation. Wait, let's check the arithmetic again:

\(8(t + 2)+(t - 2)=1\)

\(8t+16+t - 2=1\)

\(9t + 14 = 1\)

\(9t=1 - 14=-13\)

\(t =-\frac{13}{9}\approx - 1.444\), which is not 2 or - 2, so it is in the domain. So the solution for \(t\) is \(-\frac{13}{9}\), and there are no extraneous solutions. But the user's problem may have a different intention. Wait, maybe I made a mistake in the sign when multiplying. Wait, no, \(\frac{8}{t - 2}\times(t - 2)(t + 2)=8(t + 2)\) is correct, \(\frac{1}{t + 2}\times(t - 2)(t + 2)=(t - 2)\) is correct, and \(\frac{1}{(t - 2)(t + 2)}\times(t - 2)(t + 2)=1\) is correct. Then the equation \(8(t + 2)+(t - 2)=1\) is correct. Then solving it gives \(t =-\frac{13}{9}\), which is valid. So the extraneous solutions are None.

Answer:

None