Question

use the following image to answer the question.
\\(\frac{2}{x + 5} + \frac{1}{x^2 + 4x - 5}\\)
question 36
add the following expressions and simplify the result.
select one
\\(\circ \frac{3}{x - 1}\\)
\\(\circ \frac{3}{x^2 + 4x - 5}\\)
\\(\circ \frac{2x - 1}{x^2 + 4x - 5}\\)
\\(\circ \frac{2x}{x + 5}\\)

Explanation:

Step1: Factor the denominator

Factor \(x^2 + 4x - 5\). We need two numbers that multiply to \(-5\) and add to \(4\). Those numbers are \(5\) and \(-1\). So, \(x^2 + 4x - 5=(x + 5)(x - 1)\).

Step2: Find a common denominator

The first fraction is \(\frac{2}{x + 5}\) and the second is \(\frac{1}{(x + 5)(x - 1)}\). The common denominator is \((x + 5)(x - 1)\). Rewrite the first fraction: \(\frac{2}{x + 5}=\frac{2(x - 1)}{(x + 5)(x - 1)}\).

Step3: Add the fractions

Now add \(\frac{2(x - 1)}{(x + 5)(x - 1)}+\frac{1}{(x + 5)(x - 1)}\). Combine the numerators: \(2(x - 1)+1 = 2x-2 + 1=2x - 1\)? Wait, no, wait. Wait, no, let's recalculate. Wait, \(2(x - 1)=2x-2\), then \(2x - 2+1 = 2x - 1\)? But wait, maybe I made a mistake. Wait, no, wait the options. Wait, no, wait let's check again. Wait, no, the first fraction: \(\frac{2}{x + 5}\), common denominator \((x + 5)(x - 1)\), so multiply numerator and denominator by \((x - 1)\): \(\frac{2(x - 1)}{(x + 5)(x - 1)}\). Then add \(\frac{1}{(x + 5)(x - 1)}\): \(\frac{2(x - 1)+1}{(x + 5)(x - 1)}=\frac{2x-2 + 1}{(x + 5)(x - 1)}=\frac{2x - 1}{(x + 5)(x - 1)}\)? But \((x + 5)(x - 1)=x^2 + 4x - 5\), so that's \(\frac{2x - 1}{x^2 + 4x - 5}\)? Wait, but wait, maybe I messed up. Wait, no, wait the options. Wait, no, wait let's check the options again. Wait, option C is \(\frac{2x - 1}{x^2 + 4x - 5}\), but wait, maybe I made a mistake in factoring? Wait, no, \(x^2 + 4x - 5=(x + 5)(x - 1)\) is correct. Wait, but let's check the addition again. Wait, \(\frac{2}{x + 5}+\frac{1}{x^2 + 4x - 5}=\frac{2(x - 1)+1}{(x + 5)(x - 1)}=\frac{2x-2 + 1}{(x + 5)(x - 1)}=\frac{2x - 1}{(x + 5)(x - 1)}=\frac{2x - 1}{x^2 + 4x - 5}\). Wait, but let's check the options. Option C is \(\frac{2x - 1}{x^2 + 4x - 5}\), but wait, maybe I made a mistake. Wait, no, wait another way. Wait, maybe the first step was wrong. Wait, no, let's check the answer options again. Wait, option A is \(\frac{3}{x - 1}\). Wait, how? Wait, maybe I factored wrong? Wait, no, \(x^2 + 4x - 5=(x + 5)(x - 1)\). Wait, let's try adding again. Wait, \(\frac{2}{x + 5}+\frac{1}{(x + 5)(x - 1)}\). Let's factor out \(\frac{1}{x + 5}\): \(\frac{1}{x + 5}(2+\frac{1}{x - 1})=\frac{1}{x + 5}(\frac{2(x - 1)+1}{x - 1})=\frac{1}{x + 5}(\frac{2x-2 + 1}{x - 1})=\frac{1}{x + 5}(\frac{2x - 1}{x - 1})=\frac{2x - 1}{(x + 5)(x - 1)}\). But that's the same as before. Wait, but option A is \(\frac{3}{x - 1}\). Wait, maybe I made a mistake in the numerator. Wait, wait, \(2(x - 1)+1\): if \(x - 1\) is a factor, maybe I miscalculated. Wait, no, \(2(x - 1)=2x - 2\), plus 1 is \(2x - 1\). But maybe the problem is different. Wait, wait, maybe the original problem is \(\frac{2}{x + 5}+\frac{1}{x^2 + 4x - 5}\). Wait, let's check the options again. Wait, option A: \(\frac{3}{x - 1}\). Let's see, if we have \(\frac{2}{x + 5}+\frac{1}{(x + 5)(x - 1)}=\frac{2(x - 1)+1}{(x + 5)(x - 1)}=\frac{2x-2 + 1}{(x + 5)(x - 1)}=\frac{2x - 1}{(x + 5)(x - 1)}\). But that's not option A. Wait, maybe I made a mistake in factoring. Wait, no, \(x^2 + 4x - 5=(x + 5)(x - 1)\) is correct. Wait, maybe the numerator is \(2(x - 1)+1 = 2x - 2 + 1=2x - 1\), but option C is \(\frac{2x - 1}{x^2 + 4x - 5}\), which is the same as \(\frac{2x - 1}{(x + 5)(x - 1)}\). But wait, maybe the problem was written incorrectly? Or maybe I made a mistake. Wait, let's try another approach. Let's plug in a value for \(x\), say \(x = 2\). Then the original expression: \(\frac{2}{2 + 5}+\frac{1}{4 + 8 - 5}=\frac{2}{7}+\frac{1}{7}=\frac{3}{7}\). Now check the options:

- Option A: \(\frac{3}{2 - 1}=3\). No, that's not \(\frac{3}{7}\).
- Option B: \(\frac{3}{4 + 8 - 5}=\frac{3}{7}\). Wait, that's \(\frac{3}{7}\). Wait, what? Wait, \(x = 2\), option B: \(\frac{3}{x^2 + 4x - 5}=\frac{3}{7}\), which matches. Wait, so where did I go wrong?

Wait, let's recalculate the numerator. Wait, \(\frac{2}{x + 5}+\frac{1}{(x + 5)(x - 1)}\). Let's factor out \(\frac{1}{x + 5}\): \(\frac{1}{x + 5}(2+\frac{1}{x - 1})=\frac{1}{x + 5}(\frac{2(x - 1)+1}{x - 1})=\frac{1}{x + 5}(\frac{2x - 2 + 1}{x - 1})=\frac{1}{x + 5}(\frac{2x - 1}{x - 1})\). But if \(x = 2\), that's \(\frac{1}{7}(\frac{3}{1})=\frac{3}{7}\), which is the same as option B: \(\frac{3}{x^2 + 4x - 5}\) when \(x = 2\) is \(\frac{3}{7}\). Wait, so my previous calculation was wrong. Wait, \(2(x - 1)+1\): if \(x - 1 = 1\) (when \(x = 2\)), then \(2(1)+1 = 3\). Oh! Wait, I see my mistake. \(2(x - 1)+1\) is \(2x - 2 + 1\), but when \(x = 2\), that's \(4 - 2 + 1 = 3\), not \(2x - 1\). Wait, \(2x - 1\) when \(x = 2\) is \(3\), which is correct. Wait, \(2x - 1 = 3\) when \(x = 2\). So \(\frac{2x - 1}{(x + 5)(x - 1)}\) when \(x = 2\) is \(\frac{3}{7}\), which is the same as option B: \(\frac{3}{x^2 + 4x - 5}\) (since \(x^2 + 4x - 5 = 7\) when \(x = 2\)). Wait, so \(\frac{2x - 1}{(x + 5)(x - 1)}=\frac{2x - 1}{x^2 + 4x - 5}\), but when \(x = 2\), that's \(\frac{3}{7}\), and option B is \(\frac{3}{x^2 + 4x - 5}\), which is also \(\frac{3}{7}\) when \(x = 2\). Wait, that means \(2x - 1 = 3\) when \(x = 2\), so \(2x - 1 = 3\) implies \(x = 2\), which is consistent. Wait, but how? Wait, no, \(2(x - 1)+1 = 2x - 2 + 1 = 2x - 1\), which is correct. But when \(x = 2\), \(2x - 1 = 3\), so the numerator is 3, and the denominator is \((x + 5)(x - 1)=7\), so \(\frac{3}{7}\), which is the same as option B: \(\frac{3}{x^2 + 4x - 5}\) (since \(x^2 + 4x - 5 = 7\) when \(x = 2\)). Wait, so \(\frac{2x - 1}{(x + 5)(x - 1)}=\frac{3}{x - 1}\)? No, wait, \(\frac{2x - 1}{(x + 5)(x - 1)}\) can be simplified? Wait, no, unless \(2x - 1\) has a factor of \(x + 5\), which it doesn't. Wait, but when \(x = 2\), both option B and option C give \(\frac{3}{7}\). Wait, let's check \(x = 0\). Original expression: \(\frac{2}{5}+\frac{1}{-5}=\frac{2}{5}-\frac{1}{5}=\frac{1}{5}\). Option B: \(\frac{3}{0 + 0 - 5}=\frac{3}{-5}=-\frac{3}{5}\). No, that's not \(\frac{1}{5}\). Option C: \(\frac{0 - 1}{0 + 0 - 5}=\frac{-1}{-5}=\frac{1}{5}\). Ah! There we go. So when \(x = 0\), original expression is \(\frac{2}{5}+\frac{1}{-5}=\frac{1}{5}\). Option C: \(\frac{2(0)-1}{0 + 0 - 5}=\frac{-1}{-5}=\frac{1}{5}\). Option B: \(\frac{3}{-5}=-\frac{3}{5}\). So my mistake earlier was in the \(x = 2\) calculation. Wait, \(x = 0\): original expression is \(\frac{2}{0 + 5}+\frac{1}{0 + 0 - 5}=\frac{2}{5}-\frac{1}{5}=\frac{1}{5}\). Option C: \(\frac{2(0)-1}{0 + 0 - 5}=\frac{-1}{-5}=\frac{1}{5}\). Option B: \(\frac{3}{-5}=-\frac{3}{5}\). So option C is correct. Wait, so where did I go wrong before? When \(x = 2\), original expression: \(\frac{2}{7}+\frac{1}{7}=\frac{3}{7}\). Option C: \(\frac{2(2)-1}{7}=\frac{3}{7}\), which is correct. Option B: \(\frac{3}{7}\), which is also correct? Wait, no, when \(x = 0\), option B is wrong. So my initial calculation was correct: \(\frac{2(x - 1)+1}{(x + 5)(x - 1)}=\frac{2x - 2 + 1}{(x + 5)(x - 1)}=\frac{2x - 1}{(x + 5)(x - 1)}=\frac{2x - 1}{x^2 + 4x - 5}\), which is option C. So the correct answer is option C: \(\frac{2x - 1}{x^2 + 4x - 5}\). Wait, but when \(x = 2\), that's \(\frac{3}{7}\), which matches. When \(x = 0\), that's \(\frac{-1}{-5}=\frac{1}{5}\), which matches the original expression. So the correct answer is option C.

Answer:

C. \(\frac{2x - 1}{x^2 + 4x - 5}\)