Question

1. \\(\frac{k + 5}{k - 1} - \frac{1}{k^2 - 8k + 7} = 1\\)

Explanation:

Step1: Factor the denominator

First, factor the quadratic in the denominator: \(k^2 - 8k + 7=(k - 1)(k - 7)\). The equation is \(\frac{k + 5}{k - 1}-\frac{1}{(k - 1)(k - 7)} = 1\).

Step2: Multiply through by the LCD

The least common denominator (LCD) of the fractions is \((k - 1)(k - 7)\). Multiply each term by the LCD:
\[ \begin{align*} (k + 5)(k - 7)-1&=(k - 1)(k - 7)\\ k^2-7k + 5k-35 - 1&=k^2-7k - k + 7\\ k^2-2k-36&=k^2-8k + 7 \end{align*} \]

Step3: Simplify the equation

Subtract \(k^2\) from both sides: \(-2k-36=-8k + 7\). Add \(8k\) to both sides: \(6k-36 = 7\). Add 36 to both sides: \(6k=43\)? Wait, no, wait, let's check the expansion again. Wait, \((k + 5)(k - 7)=k^2-7k+5k - 35=k^2 - 2k-35\), then subtract 1: \(k^2-2k - 36\). The right side: \((k - 1)(k - 7)=k^2-7k - k + 7=k^2-8k + 7\). So moving all terms to left: \(k^2-2k-36-(k^2-8k + 7)=0\) \(\Rightarrow k^2-2k-36 - k^2 + 8k-7 = 0\) \(\Rightarrow 6k-43 = 0\)? Wait, that can't be. Wait, maybe I made a mistake in the original equation. Wait, the original equation is \(\frac{k + 5}{k - 1}-\frac{1}{k^2-8k + 7}=1\), and \(k^2-8k + 7=(k - 1)(k - 7)\). So multiplying by \((k - 1)(k - 7)\) (note \(k eq1,k eq7\)):

\((k + 5)(k - 7)-1=(k - 1)(k - 7)\)

Expand \((k + 5)(k - 7)=k^2-7k + 5k-35=k^2-2k-35\), then subtract 1: \(k^2-2k-36\)

Right side: \(k^2-8k + 7\)

So \(k^2-2k-36=k^2-8k + 7\)

Subtract \(k^2\) from both sides: \(-2k-36=-8k + 7\)

Add \(8k\) to both sides: \(6k-36 = 7\)

Add 36 to both sides: \(6k=43\)? Wait, that gives \(k=\frac{43}{6}\), but that seems odd. Wait, maybe the original equation was written wrong? Wait, maybe the first fraction is \(\frac{k + 5}{k - 1}\) and the second is \(\frac{1}{k^2-8k + 7}\), but maybe the numerator of the second fraction is different? Wait, no, the user's image shows \(\frac{k + 5}{k - 1}-\frac{1}{k^2-8k + 7}=1\). Wait, maybe I miscalculated the expansion. Let's do it again:

Left side after multiplying by LCD: \((k + 5)(k - 7)-1\)

\((k + 5)(k - 7)=k^2-7k+5k - 35=k^2-2k-35\)

Subtract 1: \(k^2-2k-36\)

Right side: \((k - 1)(k - 7)=k^2-8k + 7\)

So \(k^2-2k-36=k^2-8k + 7\)

Subtract \(k^2\): \(-2k-36=-8k + 7\)

Add \(8k\): \(6k-36=7\)

Add 36: \(6k=43\)

\(k=\frac{43}{6}\approx7.166\), but we have to check for extraneous solutions. The denominators are \(k - 1\) and \((k - 1)(k - 7)\), so \(k eq1\) and \(k eq7\). \(\frac{43}{6}\approx7.166\) is not 1 or 7, so it's valid? Wait, but maybe the original equation was \(\frac{k + 5}{k - 1}-\frac{1}{k^2-8k + 7}=1\), but maybe I misread the numerator. Wait, maybe the first fraction is \(\frac{k + 5}{k - 1}\) and the second is \(\frac{1}{k^2-8k + 7}\), but let's check again. Wait, maybe the equation is \(\frac{k + 5}{k - 1}-\frac{1}{k^2-8k + 7}=1\), and when we solve, we get \(k=\frac{43}{6}\)? But that seems unusual. Wait, maybe there was a typo in the problem. Alternatively, maybe the original equation is \(\frac{k + 5}{k - 1}-\frac{1}{k^2-8k + 7}=1\), and the solution is \(k=\frac{43}{6}\). But let's check the calculation again.

Wait, let's start over.

Original equation: \(\frac{k + 5}{k - 1}-\frac{1}{k^2-8k + 7}=1\)

Factor denominator: \(k^2-8k + 7=(k - 1)(k - 7)\)

Multiply both sides by \((k - 1)(k - 7)\) ( \(k eq1,k eq7\) ):

\((k + 5)(k - 7)-1=(k - 1)(k - 7)\)

Expand left side: \(k^2-7k + 5k-35 - 1=k^2-2k-36\)

Expand right side: \(k^2-7k - k + 7=k^2-8k + 7\)

Set equal: \(k^2-2k-36=k^2-8k + 7\)

Subtract \(k^2\) from both sides: \(-2k-36=-8k + 7\)

Add \(8k\) to both sides: \(6k-36=7\)

Add 36 to both sides: \(6k=43\)

Divide by 6: \(k=\frac{43}{6}\)

Wait, but maybe the original problem was \(\frac{k + 5}{k - 1}-\frac{1}{k^2-8k + 7}=1\), and that's the solution. But let's check if when we plug back in, does it work?

Left side: \(\frac{\frac{43}{6}+5}{\frac{43}{6}-1}-\frac{1}{(\frac{43}{6})^2-8(\frac{43}{6})+7}\)

First term: \(\frac{\frac{43 + 30}{6}}{\frac{43 - 6}{6}}=\frac{73}{37}\)

Second term denominator: \(\frac{1849}{36}-\frac{344}{6}+7=\frac{1849}{36}-\frac{2064}{36}+\frac{252}{36}=\frac{1849 - 2064 + 252}{36}=\frac{37}{36}\)

So second term: \(\frac{1}{\frac{37}{36}}=\frac{36}{37}\)

So left side: \(\frac{73}{37}-\frac{36}{37}=\frac{37}{37}=1\), which matches the right side. So the solution is \(k = \frac{43}{6}\). Wait, but maybe I made a mistake in the initial expansion. Wait, no, the check works. So the solution is \(k=\frac{43}{6}\).

Wait, but maybe the original problem was written with a different numerator? For example, maybe the second fraction is \(\frac{1}{k^2-8k + 7}\) and the first is \(\frac{k + 5}{k - 1}\), and the solution is \(k=\frac{43}{6}\).

Answer:

\(k=\frac{43}{6}\)