Question

solve for t.
\frac{4}{t - 2} = \frac{1}{t - 8}
there may be 1 or 2 solutions.
t = \square or t = \square

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{4}{t - 2}=\frac{1}{t - 8}\), we cross - multiply. Cross - multiplying gives us \(4\times(t - 8)=1\times(t - 2)\).

Step2: Expand both sides

Expanding the left - hand side: \(4t-32\), and the right - hand side: \(t - 2\). So the equation becomes \(4t-32=t - 2\).

Step3: Subtract t from both sides

Subtracting \(t\) from both sides of the equation \(4t-32=t - 2\), we get \(4t-t-32=t - t-2\), which simplifies to \(3t-32=-2\).

Step4: Add 32 to both sides

Adding 32 to both sides of the equation \(3t-32=-2\), we have \(3t-32 + 32=-2 + 32\), so \(3t=30\).

Step5: Divide by 3

Dividing both sides of the equation \(3t = 30\) by 3, we get \(t=\frac{30}{3}=10\). We also need to check for extraneous solutions. The original equation has denominators \(t - 2\) and \(t - 8\). When \(t = 10\), \(t-2=10 - 2 = 8
eq0\) and \(t - 8=10 - 8 = 2
eq0\). Let's check if there was a mistake in assuming only one solution. Let's go back to the cross - multiplied equation \(4(t - 8)=t - 2\). If we made a mistake in cross - multiplying, but cross - multiplication is valid for non - zero denominators. Let's re - examine the steps. The equation \(4t-32=t - 2\) is a linear equation, which has only one solution. Wait, maybe I made a mistake in the problem interpretation. Wait, the original equation is \(\frac{4}{t - 2}=\frac{1}{t - 8}\). Wait, maybe the user made a typo? Or maybe I misread the equation. Wait, if the equation was \(\frac{4}{t - 2}=\frac{1}{t - 8}\), it's a linear equation after cross - multiplying, so only one solution. But the problem says "there may be 1 or 2 solutions". Wait, maybe the equation is \(\frac{4}{t - 2}=\frac{1}{t - 8}\) or maybe a quadratic? Wait, no, cross - multiplying gives a linear equation. Wait, let's check again. \(4(t - 8)=t - 2\) => \(4t-32=t - 2\) => \(3t=30\) => \(t = 10\). There is only one solution. Maybe the problem was supposed to be a rational equation that leads to a quadratic? For example, if the equation was \(\frac{4}{t - 2}=\frac{1}{t - 8}+ \text{something}\), but as per the given equation, it's a linear equation after cross - multiplying. So the solution is \(t = 10\). But the problem has two boxes, maybe a mistake in the problem statement. But based on the given equation \(\frac{4}{t - 2}=\frac{1}{t - 8}\), the solution is \(t = 10\).

Answer:

\(t = 10\) (Note: If there was a mistake in the problem and the equation was quadratic, for example, if the equation was \(\frac{4}{t - 2}=\frac{1}{t - 8}\) with a different setup, but based on the given equation, only one solution. However, if we assume that the equation was \(\frac{4}{t - 2}=\frac{1}{t - 8}\) and there was a miscalculation, but our steps are correct. So the answer is \(t = 10\))