Question

simplify the expression: \\(\frac{x + 6}{x^2 + 3x} - \frac{5}{x^2 + x - 6}\\)\
(1 point)\
\\(\bigcirc\\) \\(\frac{x - 4}{x^2 - 2x}\\)\
\\(\bigcirc\\) \\(\frac{x - 1}{x + 6}\\)\
\\(\bigcirc\\) \\(\frac{x - 1}{x^3 + 3x^2 - 2x}\\)\
\\(\bigcirc\\) \\(\frac{x^2 + 9x - 12}{x^3 + 3x^2 - 2x}\\)

Explanation:

Step1: Factor denominators

Factor \(x^2 + 3x\) as \(x(x + 3)\), and factor \(x^2 + x - 6\) as \((x + 3)(x - 2)\). So the expression becomes \(\frac{x + 6}{x(x + 3)}-\frac{5}{(x + 3)(x - 2)}\).

Step2: Find common denominator

The common denominator is \(x(x + 3)(x - 2)\). Rewrite each fraction:
\(\frac{(x + 6)(x - 2)}{x(x + 3)(x - 2)}-\frac{5x}{x(x + 3)(x - 2)}\)

Step3: Expand numerators

Expand \((x + 6)(x - 2)\) to \(x^2 + 4x - 12\), and the second numerator is \(5x\). So we have \(\frac{x^2 + 4x - 12 - 5x}{x(x + 3)(x - 2)}\).

Step4: Simplify numerator

Simplify the numerator: \(x^2 + 4x - 12 - 5x=x^2 - x - 12\)? Wait, no, wait. Wait, let's recalculate: \((x + 6)(x - 2)=x^2 - 2x + 6x - 12=x^2 + 4x - 12\). Then subtract \(5x\): \(x^2 + 4x - 12 - 5x=x^2 - x - 12\)? Wait, that's not right. Wait, maybe I made a mistake. Wait, no, let's check again. Wait, the original problem is \(\frac{x + 6}{x^2 + 3x}-\frac{5}{x^2 + x - 6}\). Wait, maybe I messed up the factoring. Wait, \(x^2 + x - 6=(x + 3)(x - 2)\), correct. \(x^2 + 3x=x(x + 3)\), correct. Then common denominator is \(x(x + 3)(x - 2)\). Then first fraction: \(\frac{(x + 6)(x - 2)}{x(x + 3)(x - 2)}\), second fraction: \(\frac{5x}{x(x + 3)(x - 2)}\). Now expand \((x + 6)(x - 2)\): \(x^2 - 2x + 6x - 12=x^2 + 4x - 12\). Then subtract \(5x\): \(x^2 + 4x - 12 - 5x=x^2 - x - 12\)? Wait, no, that can't be. Wait, maybe the answer options are different. Wait, let's try another approach. Wait, maybe I made a mistake in the problem. Wait, the options have \(\frac{x - 1}{x^3 + 3x^2 - 2x}\). Let's factor the denominator of the option: \(x^3 + 3x^2 - 2x=x(x^2 + 3x - 2)\)? No, wait, \(x^3 + 3x^2 - 2x=x(x^2 + 3x - 2)\)? No, wait, maybe the common denominator is \(x(x + 3)(x - 2)=x^3 + 3x^2 - 2x - 6x\)? Wait, no, \(x(x + 3)(x - 2)=x[(x + 3)(x - 2)]=x(x^2 + x - 6)=x^3 + x^2 - 6x\). Wait, I think I messed up the factoring of the denominator in the option. Wait, the third option is \(\frac{x - 1}{x^3 + 3x^2 - 2x}\). Let's factor the denominator: \(x^3 + 3x^2 - 2x=x(x^2 + 3x - 2)\)? No, that's not right. Wait, maybe the original problem's denominators were factored wrong. Wait, let's start over.

Wait, the first denominator: \(x^2 + 3x=x(x + 3)\). Second denominator: \(x^2 + x - 6\). Let's factor \(x^2 + x - 6\): looking for two numbers that multiply to -6 and add to 1. Those numbers are 3 and -2. So \((x + 3)(x - 2)\). Correct. So the expression is \(\frac{x + 6}{x(x + 3)}-\frac{5}{(x + 3)(x - 2)}\). The common denominator is \(x(x + 3)(x - 2)\). So rewrite each fraction:

First fraction: \(\frac{(x + 6)(x - 2)}{x(x + 3)(x - 2)}\)

Second fraction: \(\frac{5x}{x(x + 3)(x - 2)}\)

Now subtract the numerators: \((x + 6)(x - 2)-5x\)

Expand \((x + 6)(x - 2)\): \(x^2 - 2x + 6x - 12=x^2 + 4x - 12\)

Subtract \(5x\): \(x^2 + 4x - 12 - 5x=x^2 - x - 12\)? Wait, that's not matching the options. Wait, maybe the problem was written incorrectly? Wait, no, the options have \(\frac{x - 1}{x^3 + 3x^2 - 2x}\). Let's check the numerator again. Wait, maybe I made a mistake in the sign. Wait, the second fraction is subtracted, so it's \((x + 6)(x - 2)-5x\). Wait, maybe the original problem is \(\frac{x + 6}{x^2 + 3x}-\frac{5}{x^2 - x - 6}\)? No, the problem says \(x^2 + x - 6\). Wait, let's check the third option's denominator: \(x^3 + 3x^2 - 2x=x(x^2 + 3x - 2)\)? No, that's not. Wait, maybe the common denominator is \(x(x + 3)(x - 2)=x^3 + x^2 - 6x\). But the third option's denominator is \(x^3 + 3x^2 - 2x\). Hmm. Wait, maybe I made a mistake in factoring the second denominator. Wait, \(x^2 + x - 6\): discriminant is \(1 + 24 = 25\), square root of 25 is 5. So roots are \(\frac{-1 \pm 5}{2}\), so \(x = 2\) and \(x = -3\). So \((x + 3)(x - 2)\). Correct. So that's right. Wait, maybe the numerator simplifies to \(x - 1\). Wait, let's try plugging in a value for \(x\). Let's take \(x = 1\). Then original expression: \(\frac{1 + 6}{1 + 3}-\frac{5}{1 + 1 - 6}=\frac{7}{4}-\frac{5}{-4}=\frac{7}{4}+\frac{5}{4}=\frac{12}{4}=3\). Now check the options:

First option: \(\frac{1 - 4}{1 - 2}=\frac{-3}{-1}=3\). Wait, that's 3. Wait, but let's check with \(x = 1\) in first option: \(\frac{1 - 4}{1 - 2}=\frac{-3}{-1}=3\). Original expression at \(x = 1\) is 3. Let's check \(x = 4\). Original expression: \(\frac{4 + 6}{16 + 12}-\frac{5}{16 + 4 - 6}=\frac{10}{28}-\frac{5}{14}=\frac{10}{28}-\frac{10}{28}=0\). First option: \(\frac{4 - 4}{16 - 8}=\frac{0}{8}=0\). So that works. Wait, but earlier algebra was wrong. Wait, let's redo the algebra.

Wait, original expression: \(\frac{x + 6}{x(x + 3)}-\frac{5}{(x + 3)(x - 2)}\)

Common denominator: \(x(x + 3)(x - 2)\)

First fraction: \(\frac{(x + 6)(x - 2)}{x(x + 3)(x - 2)}\)

Second fraction: \(\frac{5x}{x(x + 3)(x - 2)}\)

Subtract numerators: \((x + 6)(x - 2)-5x\)

Expand: \(x^2 - 2x + 6x - 12 - 5x=x^2 - x - 12\)? Wait, but when \(x = 1\), that's \(1 - 1 - 12=-12\), but the first option at \(x = 1\) is 3. So there's a mistake here. Wait, maybe the second denominator is \(x^2 - x - 6\) instead of \(x^2 + x - 6\)? Let's check: \(x^2 - x - 6=(x - 3)(x + 2)\). No, that doesn't help. Wait, maybe the problem is \(\frac{x + 6}{x^2 + 3x}-\frac{5}{x^2 - x - 6}\). Then denominator is \((x - 3)(x + 2)\), no. Wait, maybe I made a mistake in the sign of the second fraction. Wait, the original problem is subtraction: \(\frac{x + 6}{x^2 + 3x}-\frac{5}{x^2 + x - 6}\). So that's correct. Wait, when \(x = 1\), original expression: \(\frac{7}{4}-\frac{5}{-4}=\frac{7}{4}+\frac{5}{4}=3\). First option: \(\frac{1 - 4}{1 - 2}=\frac{-3}{-1}=3\). So that works. Let's check the algebra again. Wait, maybe the numerator is \(x - 4\) and denominator is \(x^2 - 2x\). Let's simplify the first option: \(\frac{x - 4}{x(x - 2)}\). Let's see if our expression simplifies to that.

Wait, our expression after common denominator: \(\frac{(x + 6)(x - 2)-5x}{x(x + 3)(x - 2)}\). Wait, no, that's not. Wait, maybe the common denominator is \(x(x - 2)\)? No, because the first denominator has \(x + 3\) and the second has \(x + 3\) and \(x - 2\). Wait, maybe I made a mistake in the common denominator. Wait, no, the least common denominator for \(x(x + 3)\) and \((x + 3)(x - 2)\) is \(x(x + 3)(x - 2)\). So that's correct. Wait, let's expand the numerator again: \((x + 6)(x - 2)=x^2 + 4x - 12\), subtract \(5x\): \(x^2 - x - 12\). Factor that: \(x^2 - x - 12=(x - 4)(x + 3)\). Ah! There we go. So numerator is \((x - 4)(x + 3)\), denominator is \(x(x + 3)(x - 2)\). Cancel out \((x + 3)\) (assuming \(x eq -3\)): \(\frac{x - 4}{x(x - 2)}=\frac{x - 4}{x^2 - 2x}\). Which is the first option.

Step5: Cancel common factors

After expanding and simplifying the numerator to \((x - 4)(x + 3)\) and denominator to \(x(x + 3)(x - 2)\), we can cancel the \((x + 3)\) term (for \(x eq -3\)) to get \(\frac{x - 4}{x(x - 2)}=\frac{x - 4}{x^2 - 2x}\).

Answer:

A. \(\frac{x - 4}{x^2 - 2x}\)